Further calculus
A. Ostaszewski, J.M. Ward
MT2176, 2790176
2012
Undergraduate study in
Economics, Management,
Finance and the Social Sciences
This is an extract from a subject guide for an undergraduate course offered as part of the
University of London International Programmes in Economics, Management, Finance and
the Social Sciences. Materials for these programmes are developed by academics at the
London School of Economics and Political Science (LSE).
For more information, see: www.londoninternational.ac.uk
This guide was prepared for the University of London International Programmes by:
A. Ostaszewski, Department of Mathematics, The London School of Economics and
Political Science.
J.M. Ward, Department of Mathematics, The London School of Economics and Political
Science.
This is one of a series of subject guides published by the University. We regret that due to
pressure of work the authors are unable to enter into any correspondence relating to, or arising from, the guide. If you have any comments on this subject guide, favourable or unfavourable, please use the form at the back of this guide.
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Contents
Contents
1 Introduction
1
1.1
This subject . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.2
Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
1.3
Online study resources . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.3.1
The VLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
1.3.2
Making use of the Online Library . . . . . . . . . . . . . . . . . .
5
1.4
Using this subject guide . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
1.5
Examination advice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
1.6
The use of calculators . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
2 Limits
7
2.1
Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
2.1.1
Limits at infinity . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
2.1.2
Limits at a point . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
Some useful results that involve limits . . . . . . . . . . . . . . . . . . . .
28
2.2.1
Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
28
2.2.2
Differentiability . . . . . . . . . . . . . . . . . . . . . . . . . . . .
29
2.2.3
Taylor series and Taylor’s theorem . . . . . . . . . . . . . . . . .
31
2.2.4
L’Hôpital’s rule . . . . . . . . . . . . . . . . . . . . . . . . . . . .
34
Learning outcomes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38
Solutions to activities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
46
Solutions to exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
47
2.2
3 The Riemann integral
3.1
53
The Riemann integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
53
3.1.1
Lower and upper estimates of an area . . . . . . . . . . . . . . . .
54
3.1.2
Getting better lower and upper estimates . . . . . . . . . . . . . .
60
3.1.3
The definition of the Riemann integral . . . . . . . . . . . . . . .
63
3.1.4
What happens if the integrand isn’t continuous? . . . . . . . . . .
66
3.1.5
Some properties of the Riemann integral . . . . . . . . . . . . . .
71
i
Contents
3.2
The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . .
72
3.2.1
Motivating the FTC . . . . . . . . . . . . . . . . . . . . . . . . .
72
3.2.2
Notation: Dummy variables . . . . . . . . . . . . . . . . . . . . .
75
3.2.3
The relationship between integration and differentiation . . . . . .
76
3.2.4
Some applications of the FTC . . . . . . . . . . . . . . . . . . . .
77
3.2.5
An extension of the FTC . . . . . . . . . . . . . . . . . . . . . . .
79
Learning outcomes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
82
Solutions to activities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
82
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
93
Solutions to exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
94
4 Improper integrals
4.1
Improper integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
99
4.1.1
Improper integrals of two kinds, and a third kind . . . . . . . . .
100
4.1.2
Some further thoughts on improper integrals . . . . . . . . . . . .
102
Tests for convergence and divergence . . . . . . . . . . . . . . . . . . . .
106
4.2.1
The Direct Comparison Test . . . . . . . . . . . . . . . . . . . . .
107
4.2.2
The Limit Comparison Test . . . . . . . . . . . . . . . . . . . . .
111
4.2.3
Variable sign integrands . . . . . . . . . . . . . . . . . . . . . . .
121
Learning outcomes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
123
Solutions to activities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
123
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
129
Solutions to exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
130
4.2
5 Double integrals
5.1
ii
99
135
Double integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
135
5.1.1
Volumes over rectangular bases . . . . . . . . . . . . . . . . . . .
136
5.1.2
Defining double integrals in terms of volumes . . . . . . . . . . .
137
5.1.3
Motivating Fubini’s theorem . . . . . . . . . . . . . . . . . . . . .
139
5.1.4
Fubini’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . .
141
5.1.5
Volumes over other bases . . . . . . . . . . . . . . . . . . . . . . .
143
5.2
Change of variable techniques . . . . . . . . . . . . . . . . . . . . . . . .
152
5.3
Improper double integrals . . . . . . . . . . . . . . . . . . . . . . . . . .
162
Learning outcomes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
165
Solutions to activities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
165
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
179
Contents
Solutions to exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6 Manipulation of integrals
6.1
180
187
The manipulation of proper integrals . . . . . . . . . . . . . . . . . . . .
187
6.1.1
Joint continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . .
188
6.1.2
The manipulation rules for proper integrals . . . . . . . . . . . . .
191
6.1.3
Applications of the rules for manipulating proper integrals . . . .
194
The manipulation of improper integrals . . . . . . . . . . . . . . . . . . .
197
6.2.1
Dominated convergence
. . . . . . . . . . . . . . . . . . . . . . .
198
6.2.2
The manipulation rules for improper integrals . . . . . . . . . . .
205
6.2.3
Using the rules for manipulating improper integrals . . . . . . . .
206
Learning outcomes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
210
Solutions to activities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
210
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
216
Solutions to exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
217
6.2
7 Laplace transforms
7.1
221
What is a Laplace transform? . . . . . . . . . . . . . . . . . . . . . . . .
221
7.1.1
Some properties of the Laplace transform . . . . . . . . . . . . . .
223
7.1.2
Extending our view of Laplace transforms . . . . . . . . . . . . .
230
Using Laplace transforms . . . . . . . . . . . . . . . . . . . . . . . . . . .
233
7.2.1
Solving ODEs with constant coefficients . . . . . . . . . . . . . .
233
7.2.2
Convolutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
236
Learning outcomes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
241
Solutions to activities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
242
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
251
Solutions to exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
252
7.2
A Sample examination paper
257
B Solutions to the sample examination paper
261
iii
Contents
iv
1
Chapter 1
Introduction
In this very brief introduction, we aim to give you an idea of the nature of this subject
and to advise you on how best to approach it. We give general information about the
contents and use of this subject guide, and on recommended reading and how to use the
textbooks.
1.1
This subject
Calculus, as studied in this 200 course, is primarily the study of integrals of functions of
one and two variables.
Our approach here is not just to help you acquire proficiency in techniques and
methods, but also to help you understand some of the theoretical ideas behind these.
For example, after completing this course, you will hopefully understand how certain
kinds of definite integral are defined and how to deal with integrals where the integrand
is a function of two variables.
Aims of the course
The broad aims of this course are:
to enable students to acquire skills in the methods of calculus, as required for their
use in further mathematics subjects and economics-based subjects;
to prepare students for further courses in mathematics and/or related disciplines.
However, as emphasised above, we do also want you to understand why certain methods
work: this is one of the ‘skills’ that you should acquire. Indeed, the examination will not
simply test your ability to perform routine calculations, it will also probe your
knowledge and understanding of the principles that underlie the material.
Learning outcomes
We now state the broad learning outcomes of this course, as a whole. At the end of this
course and having completed the essential reading and activities, you should be able to:
1
1
1. Introduction
demonstrate knowledge of the subject matter, terminology, techniques and
conventions covered in the subject;
demonstrate an understanding of the underlying principles of the subject;
demonstrate the ability to solve problems involving an understanding of the
concepts.
There are a couple of things that we should stress at this point. Firstly, note the
intention that you will be able to solve unseen problems. This means simply that you
will be expected to be able to use your knowledge and understanding of the material to
solve problems that are not completely standard. This is not something you should
worry unduly about: all courses in mathematics expect this, and you will never be
expected to do anything that cannot be done using the material of this course.
Secondly, we expect you to be able to ‘demonstrate knowledge and understanding’ and
you might well wonder how you would demonstrate this in the examination. Well, it is
precisely by being able to grapple successfully with unseen, non-routine, questions that
you will indicate that you have a proper understanding of the topic.
Topics covered
Descriptions of the topics to be covered appear in the relevant chapters. However, it is
useful to give a brief overview at this stage.
We start by introducing the limit of a function of one variable and, in particular, how
this can be used to define what it means to say that a function is continuous. We then
introduce the Riemann integral and explain its relationship to differentiation via the
Fundamental Theorem of Calculus. This leads on to a discussion of improper integrals
and, in particular, some tests that we can use to determine whether such integrals are
convergent or divergent. We then turn our attention to functions of two variables, in
particular, how we can integrate such functions over certain regions and how we can
manipulate such integrals. We then discuss Laplace transforms and some of their
important applications.
Throughout this subject guide, the emphasis will be on the theory as much as on the
methods. That is to say, our aim in this subject is not only to provide you with some
useful techniques and methods from calculus, but to also enable you to understand why
these techniques work.
1.2
Reading
There are many books that would be useful for this subject. We recommend two in
particular, and a couple of others for additional, further reading. (You should note,
however, that there are very many books suitable for this course. Indeed, almost any
text on first-year university calculus will cover the majority of the material.)
Textbook reading is essential as textbooks will provide you with more in-depth
explanations than you will find in this subject guide, and they will also provide many
more examples to study and exercises to work through. The books listed are the ones
we have referred to in this subject guide.
2
1.3. Online study resources
Essential reading
Detailed reading references in this subject guide refer to the editions of the set
textbooks listed below. New editions of one or more of these textbooks may have been
published by the time you study this course. You can use a more recent edition of any
of the books; use the detailed chapter and section headings and the index to identify
relevant readings. Also check the virtual learning environment (VLE) regularly for
updated guidance on readings.
+ Binmore, K. and J. Davies Calculus: concepts and methods. (Cambridge:
Cambridge University Press, 2002) second revised edition [ISBN 9780521775410].
+ Ostaszewski, A. Advanced mathematical methods. (Cambridge: Cambridge
University Press, 1991) [ISBN 9780521289641].
Both of these texts, when used wisely, will provide you with a large number of examples
for you to study and exercises for you to attempt. It is recommended that you purchase
both of these.
Further reading
Once you have covered the essential reading you are then free to read around the
subject area in any text, paper or online resource. You will need to support your
learning by reading as widely as possible and by thinking about how these principles
apply in the real world. To help you read extensively, you have free access to the VLE
and University of London Online Library (see Section 1.3.2). However, two useful
textbooks that we have referred to in this subject guide are the following.
+ Adams, R.A. and C. Essex Calculus: A complete course. (Toronto: Pearson, 2009)
seventh edition [ISBN 9780321549280].
+ Wrede, R. C. and M. Spiegel Schaum’s outline of advanced calculus. (London:
McGraw-Hill, 2010) third edition [ISBN 9780071623667].
Adams and Essex (which is merely an example from a large range of very similar
calculus textbooks) is a detailed calculus textbook which contains much material which
is beyond the scope of this course. Wrede and Spiegel contains a brief summary of some
of the course material but is useful as it contains a large number of worked examples
and exercises. Both of these texts are suitable as sources of additional explanation,
examples and exercises, but they are probably not worth purchasing.
1.3
Online study resources
In addition to the subject guide and the essential reading, it is crucial that you take
advantage of the study resources that are available online for this course, including the
VLE and the Online Library.
3
1
1
1. Introduction
You can access the VLE, the Online Library and your University of London email
account via the Student Portal at
http://my.londoninternational.ac.uk
You should have received your login details for the Student Portal with your official
offer, which was emailed to the address that you gave on your application form. You
have probably already logged in to the Student Portal in order to register! As soon as
you registered, you will automatically have been granted access to the VLE, Online
Library and your fully functional University of London email account.
If you forget your login details at any point, please email [email protected]
quoting your student number.
1.3.1
The VLE
The VLE, which complements this subject guide, has been designed to enhance your
learning experience, providing additional support and a sense of community. It forms an
important part of your study experience with the University of London and you should
access it regularly.
The VLE provides a range of resources for EMFSS courses:
Self-testing activities: Doing these allows you to test your own understanding of
subject material.
Electronic study materials: The printed materials that you receive from the
University of London are available to download, including updated reading lists
and references.
Past examination papers and Examiners’ commentaries: These provide advice on
how each examination question might best be answered.
A student discussion forum: This is an open space for you to discuss interests and
experiences, seek support from your peers, work collaboratively to solve problems
and discuss subject material.
Videos: There are recorded academic introductions to the subject, interviews and
debates and, for some courses, audio-visual tutorials and conclusions.
Recorded lectures: For some courses, where appropriate, the sessions from previous
years’ Study Weekends have been recorded and made available.
Study skills: Expert advice on preparing for examinations and developing your
digital literacy skills.
Feedback forms.
Some of these resources are available for certain courses only, but we are expanding our
provision all the time and you should check the VLE regularly for updates.
4
1.4. Using this subject guide
1.3.2
Making use of the Online Library
The Online Library contains a huge array of journal articles and other resources to help
you read widely and extensively.
To access the majority of resources via the Online Library at
http://tinyurl.com/ollathens
you will either need to use your University of London Student Portal login details, or
you will be required to register and use an Athens login.
The easiest way to locate relevant content and journal articles in the Online Library is
to use the Summon search engine.
If you are having trouble finding an article listed in a reading list, try removing any
punctuation from the title, such as single quotation marks, question marks and colons.
For further advice, please see the online help pages at
www.external.shl.lon.ac.uk/summon/about.php
1.4
Using this subject guide
We have already mentioned that this subject guide is not a textbook. It is important
that you read textbooks in conjunction with the subject guide and that you try
problems from the textbooks. The exercises at the end of the main chapters of this
subject guide are a very useful resource and you should try them once you think you
have mastered the material from the chapter. You should really try these exercises
before consulting the solutions, as simply reading the solutions provided will not help
you at all. Sometimes, the solutions we provide will just be an overview of what is
required, i.e. an indication of how you should answer the questions, but in the
examination, you must always show all of your calculations. It is vital that you develop
and enhance your problem-solving skills and the only way to do this is to try lots of
exercises.
1.5
Examination advice
Important: the information and advice given here are based on the examination
structure used at the time this subject guide was written. Please note that subject
guides may be used for several years. Because of this we strongly advise you to always
check both the current Regulations for relevant information about the examination, and
the VLE where you should be advised of any forthcoming changes. You should also
carefully check the rubric/instructions on the paper you actually sit and follow those
instructions.
Remember, it is important to check the VLE for:
Up-to-date information on examination and assessment arrangements for this
course.
5
1
1
1. Introduction
Where available, past examination papers and Examiners’ commentaries for the
course which give advice on how each question might best be answered.
This course is assessed by a two hour unseen written examination. There are no
optional topics in this subject: you should study them all and this is reflected in the
structure of the examination paper. There are five questions (each worth 20 marks) and
all questions are compulsory. A sample examination paper may be found in an appendix
to this subject guide.
Please do not think that the questions in your real examination will necessarily be very
similar to the exercises in this subject guide or those in the sample examination paper.
The examination is designed to test you. You will get examination questions unlike the
questions in this subject guide. The whole point of examining is to see whether you can
apply your knowledge in familiar and unfamiliar settings. The Examiners (nice people
though they are) have an obligation to surprise you! For this reason, it is important
that you try as many examples as possible from the subject guide and from the
textbooks. This is not so that you can cover any possible type of question the
Examiners can think of! It is so that you get used to confronting unfamiliar questions,
grappling with them, and finally coming up with the solution.
Do not panic if you cannot completely solve an examination question. There are many
marks to be awarded for using the correct approach or method.
1.6
The use of calculators
You will not be permitted to use calculators of any type in the examination. This is not
something that you should worry about: the Examiners are interested in assessing that
you understand the key concepts, ideas, methods and techniques, and will set questions
which do not require the use of a calculator.
6
Chapter 2
Limits
2
Essential reading
(For full publication details, see Chapter 1.)
+ Ostaszewski (1991) Sections 17.4–17.6 and Section 18.8.
Further reading
+ Adams and Essex (2010) Sections 1.2–1.4, parts of Section 2.2, Sections 4.3 and
4.9–4.10.
+ Wrede and Spiegel (2010) parts of Chapters 3, 4 and 11.
Aims and objectives
The objectives of this chapter are:
to see what a limit is and how they can be found in a variety of different situations;
to examine the relationship between limits, continuity and differentiability.
Specific learning outcomes can be found near the end of this chapter.
2.1
Limits
We encountered the general idea behind limits in 174 Calculus and, although we used
the idea there, we never gave a thorough account of what was involved. In this section,
we will make the idea behind a limit more precise, but our account will still be fairly
informal.1 And, once we have done this, we will look at some useful results that involve
limits such as the use of limits to define what it means to say that a function is
continuous or differentiable. We will also see how our understanding of Taylor series can
be extended by using Taylor’s theorem and we will end this chapter by considering
L’Hôpital’s rule which will allow us to calculate some of the ‘trickier’ limits that we will
encounter.
1
That is, we will say enough to make the idea of a limit precise and see how to calculate limits, but
we will not give a rigorous mathematical treatment of limits like the one you will see in 116 Abstract
Mathematics.
7
2. Limits
2.1.1
2
Limits at infinity
Limits ‘at infinity’ are concerned with the behaviour of a function, f (x), as x tends to
infinity, a situation we denoted by ‘x → ∞’ in 174 Calculus. Indeed, as we saw in that
course, this kind of information is useful when we were sketching the graph of a function
because it told us what was happening as x gets ‘very large’. In this section, we consider
exactly what this kind of limit means and see how we can find such a limit (if it exists)
in some straightforward cases.
Finite limits
Suppose that l is a real number and that f (x) is a function, we start by asking what it
means to say that f (x) → l as x → ∞, i.e. what it means to say that
lim f (x) = l.
x→∞
Intuitively, based on what we saw in 174 Calculus, we would want to say that this
means that the curve y = f (x) has a horizontal asymptote given by y = l, i.e. a
horizontal line that the curve gets arbitrarily close to. But, more specifically, we mean
that however close we would like f (x) to be to l, there is some suitably large value of x,
say X, for which f (x) is as close to l as we wanted if x ≥ X.2 For instance, in Figure 2.1
we have the graphs of the two functions where f (x) → l as x → ∞.
y
y
y = f (x)
y = f (x)
l
l
x
O
(a)
x
O
(b)
Figure 2.1: Two functions for which f (x) → l as x → ∞ where l > 0 is some real number.
2
Technically, as you will see in 116 Abstract Mathematics, we say that f (x) → l as x → ∞ for some
real number, l, if
for any ε > 0, there is an X such that, for all x ≥ X, |f (x) − l| < ε.
Here, the value of ε > 0 tells us how close we want to be and, if we can show that there is an X such that
|f (x) − l| < ε for all x ≥ X, then we have established that f (x) is as close to l as we wanted for these
values of x. Indeed, if we can do this for any value of ε, we can guarantee that f (x) is getting ‘arbitrarily
close’ to l as x → ∞. However, we will not make use of this formal definition here.
8
2.1. Limits
Notice, however, that we must take some care when we describe the fact that f (x) → l
as x → ∞. In particular, we don’t want to say that f (x) → l as x → ∞ because f (x)
gets ‘closer and closer’ to l when we take larger and larger values of x. This is because,
if we consider the graph in
Figure 2.1(a), we see that f (x) also gets ‘closer and closer’ to, say, zero even
though that is clearly not its limit as x → ∞.
Figure 2.1(b), we see that at some points f (x) is ‘heading towards’ l and sometimes
it is ‘heading away’ from l and so f (x) is not always getting ‘closer and closer’ to l
even though l clearly is its limit as x → ∞.
Let’s now consider how we can actually find such limits.
Finding finite limits
Usually, we can find finite limits by considering some basic functions that have finite
limits and then, by using some appropriate rules about how limits work, we can find the
limits of certain combinations of these basic functions. So, we start by stating some
finite limits that arise from basic functions, i.e.
(a) If f (x) = a where a is a constant, then f (x) → a as x → ∞.
(b) If f (x) = 1/xb where b > 0 is a constant, then f (x) → 0 as x → ∞.
(c) If f (x) = 1/cx where c > 1 is a constant, then f (x) → 0 as x → ∞.
Then, we have the rules which tell us how to find the limits of certain combinations of
these basic functions which are stated in the following theorem.
Theorem 2.1 If f (x) → l and g(x) → m as x → ∞ where l, m ∈ R, then
(a) if c is a constant, cf (x) → cl as x → ∞;
(b) f (x) + g(x) → l + m as x → ∞;
(c) f (x)g(x) → lm as x → ∞;
(d) if m 6= 0,
l
f (x)
→
as x → ∞;
g(x)
m
(e) if b is a constant and l > 0, [f (x)]b → lb .
We will not prove this theorem here even though some of the rules may be fairly
obvious, but it is important that you treat these rules with some care. In particular, as
we require that m 6= 0 in Theorem 2.1(d), this tells us nothing about
f (x)
,
x→∞ g(x)
lim
if g(x) → 0 as x → ∞,3 but we will say more about this later. Also observe that in
Theorem 2.1(e), we require that l > 0 because, if we had b = 1/2 (say), this would make
3
As we should expect since l/m makes no sense if m = 0 because we can never divide by zero!
9
2
2. Limits
no sense in cases where l < 0 or, indeed, in some cases where l = 0 as we will see in
Activity 2.7. However, we now consider some examples of how we use the results above.
2
Example 2.1
Find lim
x→∞
1
1
3+ − x .
x 2
Using (a), (b) and (c) respectively, it should be obvious that
lim 3 = 3,
x→∞
1
= 0 and
x→∞ x
lim
1
= 0,
x→∞ 2x
lim
so, using Theorem 2.1(a) and (b) respectively, we have
1
1
lim − x = −0 = 0 and lim 3 +
= 3 + 0 = 3.
x→∞
x→∞
2
x
Then, using Theorem 2.1(b) again, this gives us
1
1
1
1
lim 3 + − x = lim
3+
+ − x
= 3 + 0 = 3,
x→∞
x→∞
x 2
x
2
as the answer.
Note: As this is fairly obvious, once you have understood the results above, you
would normally just write
1
1
lim 3 + − x = 3 + 0 − 0 = 3,
x→∞
x 2
since it is easy to find the limit of each of the three terms and hence the limit of this
combination of them.
Example 2.2
p
4 − 1/x
Find lim
.
x→∞ 1 − 4/x3
Using (b) and Theorem 2.1(a), it should be obvious that
4
1
lim −
= −0 = 0 and lim − 3 = −4(0) = 0,
x→∞
x→∞
x
x
so, using (a) and Theorem 2.1(b), we have
1
1
= lim 4 + −
= 4 + 0 = 4,
lim 4 −
x→∞
x→∞
x
x
and
4
4
lim 1 − 3 = lim 1 + − 3
1 + 0 = 1.
x→∞
x→∞
x
x
Now, the first of these limits is positive and so, using Theorem 2.1(e), we have
r
1/2
1
1
lim 4 − = lim 4 −
= 41/2 = 2,
x→∞
x x→∞
x
10
2.1. Limits
and the second of these limits is non-zero and so, using Theorem 2.1(d), we have
p
4 − 1/x
2
lim
= = 2,
3
x→∞ 1 − 4/x
1
as the answer.
Note: As this is also fairly obvious, once you have understood the results above, you
would normally just write
p
√
4 − 1/x
4−0
2
= = 2,
lim
=
3
x→∞ 1 − 4/x
1−0
1
having taken care to observe that we are taking the square root of a positive number
and that we are not dividing by zero.
Of course, given what we have seen in these two examples, it should be obvious that we
could extend Theorem 2.1 by including the two results in the next activity.
Activity 2.1 Use Theorem 2.1 to show that: If f (x) → l and g(x) → m as x → ∞
where l, m ∈ R, then
f (x) − g(x) → l − m as x → ∞,
and
cf (x) + dg(x) → cl + dm as x → ∞,
where c and d are constants.
It is also useful to note that, sometimes, it is necessary to rewrite the function we are
considering before we attempt to find the limit.
Example 2.3
x3 − 3x2 + 2
Find lim
.
x→∞
4x3 + 6x
We start by noting that, in this case, we can not simply work out the limit of this
quotient by considering the limits of the numerator and the denominator as x → ∞
because neither of these have a limit which is a real number.4
In cases such as this, we employ the useful ‘trick’ of dividing the numerator and the
denominator by the highest power of x that occurs in the quotient.5 Indeed, here,
this highest power of x is x3 and so, dividing the numerator and denominator by
this, we get
1 − (3/x) + (2/x3 )
x3 − 3x2 + 2
= lim
,
lim
x→∞
x→∞
4x3 + 6x
4 + (6/x2 )
and, we can deal with this by considering the limits of the numerator and the
denominator as x → ∞ because now, the limit of the numerator is 1 and the limit of
the denominator is 4. As such, we can see that the limit we are asked to find is 1/4.
Note: As this is fairly obvious once you understand that, in such cases, we need to
divide by the highest power of x in the quotient in order to get finite limits in its
11
2
2. Limits
numerator and denominator as x → ∞, you would normally just write
1 − (3/x) + (2/x3 )
1−0+0
1
x3 − 3x2 + 2
=
lim
=
= ,
3
2
x→∞
x→∞
4x + 6x
4 + (6/x )
4+0
4
2
lim
as it is easy to find the limit once we have rewritten the quotient in this way.
Lastly, it will sometimes be useful to appeal to the so-called ‘Sandwich theorem’ when
we are asked to find the limits of certain functions that can not be usefully analysed
using the methods above.
Theorem 2.2 (The Sandwich theorem) If, for some X ∈ R, the functions f , g and h
are related by the inequality
f (x) ≤ g(x) ≤ h(x),
for all x ≥ X and there is some l ∈ R such that
lim f (x) = l
x→∞
and
lim h(x) = l,
x→∞
then we can conclude that lim g(x) = l too.
x→∞
Of course, we will not prove this here, but we will motivate it by considering an
example where it can be usefully applied.
Example 2.4
sin x
= 0.
x→∞ x
Show that lim
As, for x > 0, we have
−1 ≤ sin x ≤ 1
=⇒
−
1
sin x
1
≤
≤ ,
x
x
x
and we also have
1
lim = 0 which means that
x→∞ x
1
lim −
= 0 as well,
x→∞
x
we can use the Sandwich theorem to conclude that
sin x
= 0,
x→∞ x
lim
as required. Of course, this is obvious if we look at the graph of this function which is
illustrated in Figure 2.2 along with the graphs of the functions ±1/x that bound it.
4
In fact, both the numerator and the denominator of this quotient ‘tend to infinity’ as x → ∞, a
situation we shall consider in more detail in a moment.
5
We could call this highest power of x the ‘dominant term’ as its behaviour will ‘determine’ the
behaviour of the function in the limit.
12
2.1. Limits
2
Figure 2.2: The dashed curve in the positive quadrant is the graph of the function 1/x
and the dashed curve in the quadrant below this is the graph of the function −1/x. The
function sinx x , whose graph is the solid line, always lies between these two curves for x > 0
and, because of this, we see that it must tend to zero as x → ∞.
Activity 2.2
Find the following limits.
(a)
x2 + x + 1
lim
,
x→∞ x + x2 + x3
(b)
(x + sin x − 2)1/2
lim √
.
x→∞
x + sin x − 2
Infinite limits
If f (x) is a function, we now want to ask what it means to say that this function tends
to infinity as x → ∞, i.e. what it means to say that f (x) → ∞ as x → ∞ or, slightly
abusing our notation,6 that
lim f (x) = ∞.
x→∞
Intuitively, based on what we saw in 174 Calculus, we want to say that this means that
the function, f (x), can take arbitrarily large values as we let x get larger and larger.
But, more specifically, we mean that however large we want f (x) to be, let’s say we
want it to be larger than some real number, M , we can find a value of x, say X, for
which f (x) is larger than M for all x ≥ X.7
6
This is an abuse of our notation since, technically, what we denote by ‘∞’ is not a real number and
so nothing can be equal to it. However, what we write here will be a very useful ‘notational convenience’.
7
Technically, as you will see in 116 Abstract Mathematics, we say that f (x) → ∞ as x → ∞ if,
for any M > 0, there is an X such that, for all x ≥ X, f (x) ≥ M .
13
2. Limits
2
We also want to consider functions, f (x), which tend to minus infinity as x → ∞, i.e.
where f (x) → −∞ as x → ∞ or, again slightly abusing our notation, where
lim f (x) = −∞.
x→∞
Of course, we saw this kind of behaviour in 174 Calculus too and we want to say that
this means that the function, f (x), can take negative values which are arbitrarily large
in magnitude as we let x get larger and larger. But, it is perhaps easier to define this in
terms of what we have just seen, i.e. we say that
lim f (x) = −∞ if lim − f (x) = ∞,
x→∞
x→∞
i.e. if −f (x) → ∞ as x → ∞, then we must have f (x) → −∞ as x → ∞. Let’s now
consider how we can actually find such limits.
Finding infinite limits
Again, as we saw above with finite limits, we can find infinite limits by considering some
basic functions that have infinite limits and then, by using some appropriate rules about
how limits work, we can then find the limits of certain combinations of these basic
functions. So, we start by stating some infinite limits that arise from basic functions, i.e.
(a) If f (x) = xb where b > 0 is a constant, then f (x) → ∞ as x → ∞.
(b) If f (x) = cx where c > 1 is a constant, then f (x) → ∞ as x → ∞.
(c) If f (x) = logd x where d > 1 is a constant, then f (x) → ∞ as x → ∞.
Then we have the rules which tell us how to find the limits of certain combinations of
these basic functions which are stated in the following theorem.
Theorem 2.3 If f (x) → ∞ as x → ∞, then
(a) if c > 0 is a constant, then cf (x) → ∞ as x → ∞.
And, if we also have g(x) → ∞ as x → ∞, then
(b) f (x) + g(x) → ∞ as x → ∞.
(c) f (x)g(x) → ∞ as x → ∞.
Whereas, if we have g(x) → m as x → ∞ where m > 0 is a real number, then
(d) f (x) + g(x) → ∞ as x → ∞.
(e) f (x)g(x) → ∞ as x → ∞.
Here, the value of M > 0 tells us how large we want f (x) to be and, if we can show that there is an
X such that f (x) ≥ M for all x ≥ X, then we have established that f (x) is always larger than M for
these values of x. Indeed, if we can do this for any value of M , we can guarantee that f (x) is getting
‘arbitrarily large’ as x → ∞. However, we will not make use of this formal definition here.
14
2.1. Limits
(f)
f (x)
→ ∞ as x → ∞.
g(x)
We will not prove this theorem here even though some of the rules may be fairly
obvious, but it is important that you treat these rules with some care. In particular,
although you can extend what we have seen in Theorem 2.3 fairly simply by doing
Activity 2.3, there are some things that we won’t be able to do at the moment as you’ll
see in Activity 2.4.
Activity 2.3 What is the analogue of Theorem 2.3(a) when c < 0 and what are the
analogues of Theorem 2.3(d)-(f) when m < 0?
Hence show that, if 0 < d < 1 is a constant, then logd x → −∞ as x → ∞.
Activity 2.4
What, if anything, can you say about the analogue of
(i) Theorem 2.3(a) when c = 0?
(ii) Theorem 2.3(b) and Theorem 2.3(c) when g(x) → −∞ as x → ∞?
(iii) Theorem 2.3(d)-(f) when m = 0?
Let’s now see how these results work by considering some examples.
Example 2.5
Find lim (x3 + x + 2).
x→∞
Using Theorem 2.3(a) and what we saw before, we have
lim x3 = ∞,
lim x = ∞ and
x→∞
x→∞
lim 2 = 2,
x→∞
so, using Theorem 2.3(d), we have
lim (x + 2) = ∞,
x→∞
so that, using Theorem 2.3(b), we get
lim (x3 + x + 2) = ∞,
x→∞
as the answer.
Example 2.6
x3 + 2x + 2
.
x→∞
x2 + 1
Find lim
We saw in Example 2.5 that the numerator of the function
x3 + 2x + 2
,
x2 + 1
tends to infinity as x → ∞ and, using similar reasoning, the denominator tends to
infinity as x → ∞ too. In particular, this means that we can’t use any of the results
15
2
2. Limits
in Theorem 2.3 on this function as it stands. However, if we divide the numerator
and the denominator of this function by the highest power of x in the denominator,8
i.e. x2 , we get
x3 + 2x + 2
x + (2/x) + (2/x2 )
=
,
x2 + 1
1 + (1/x2 )
2
and in this form the numerator still tends to infinity as x → ∞, but the denominator
now tends to one which is a positive real number. Consequently, we can use
Theorem 2.3(f) to see that
lim x + (2/x) + (2/x2 )
x3 + 2x + 2
x + (2/x) + (2/x2 )
= lim
= x→∞
lim
= ∞,
x→∞
x→∞
x2 + 1
1 + (1/x2 )
1
is the answer.
Example 2.7
x+1
.
Find lim √
x→∞
4x − 1
It should be clear that the numerator and the denominator of the function
x+1
√
,
4x − 1
both tend to infinity as x → ∞ and so we are in a similar situation to the one in
Example 2.6. So, as we did in that example, we divide the numerator and the √
denominator of this function by the highest power of x in the denominator, i.e. x,
to get
√
√
x+1
x + (1/ x)
√
= p
,
4x − 1
4 − (1/x)
and in this form the numerator still tends to infinity as x → ∞, but the denominator
now tends to two — as we saw in Example 2.2 — which is a positive real number.
Consequently, we can use Theorem 2.3(f) to see that
√
√ √
√
x
+
(1/
x)
lim
x+1
x + (1/ x)
lim √
= lim p
= x→∞
= ∞,
x→∞
2
4x − 1 x→∞
4 − (1/x)
is the answer.
Activity 2.5
x2 − sin x
.
x→∞ x + sin x
Find the limits (a) lim (x2 − x3 ) and (b) lim
x→∞
8
Of course, the highest power of x in the quotient is x3 (i.e. this is the ‘dominant term’), but if we
divide the numerator and denominator by this, we get
x3 + 2x + 2
1 + (2/x2 ) + (2/x)
=
.
2
x +1
(1/x) + (1/x3 )
And, this means that, as x → ∞, the numerator tends to one and the denominator tends to zero, a case
that we can not deal with using Theorem 2.1(d). However, we will see in Example 2.8 that we can make
sense of this once we have Theorem 2.4.
16
2.1. Limits
The relationship between infinite and finite limits
We have seen that, as x → ∞, some functions have finite limits and others have infinite
limits but now we want to briefly discuss how these two types of limit are related. The
key result here is the following theorem.
Theorem 2.4 (a) If f (x) → ∞ as x → ∞, then
1
= 0.
x→∞ f (x)
lim
(b) If f (x) → 0 as x → ∞ and there is an M ∈ R such that
f (x) > 0 for all x > M , then lim
1
= ∞.
f (x)
f (x) < 0 for all x > M , then lim
1
= −∞.
f (x)
x→∞
x→∞
Of course, the results in this theorem should be fairly obvious and we can see why if we
consider an example.
Example 2.8
Following on from Example 2.6, use Theorem 2.4 to verify that
x3 + 2x + 2
= ∞,
x→∞
x2 + 1
lim
as we found there.9
The highest power of x in the quotient is x3 ,10 and if we divide the numerator and
denominator by this, we get
1 + (2/x2 ) + (2/x3 )
2
2
1
x3 + 2x + 2
=
= 1+ 2 + 3
.
2
3
x +1
(1/x) + (1/x )
x
x (1/x) + (1/x3 )
Now, as x → ∞, the first term in this product tends to one whereas, by
Theorem 2.4, the second term tends to infinity as
1
1
+ 3 > 0,
x x
for x > 0. Consequently, by Theorem 2.3, we see that
x3 + 2x + 2
= ∞,
x→∞
x2 + 1
lim
as expected.
9
10
This example follows on from the discussion in footnote 8.
That is, x3 is the ‘dominant term’ here.
17
2
2. Limits
Limits that don’t exist
2
If we have a function f (x) and we find that
lim f (x) = c,
x→∞
where c is a real number or we find that f (x) → ∞ (or −∞) as x → ∞, we say that the
limit of f (x) as x → ∞ exists. However, not every function has a limit as x → ∞ and,
in such cases, we say that this limit does not exist.
Example 2.9
Explain why lim sin x does not exist.
x→∞
As we know from 174 Calculus, the function sin x, which is illustrated in Figure 2.3,
oscillates between the values of 1 and −1 with a period of 2π. As such, this function
has no limit as x → ∞ since it never stays arbitrarily close to any value.
Figure 2.3: The graph of the function sin x for x ≥ 0.
However, although in this case the oscillations mean that a limit doesn’t exist, we saw
in Example 2.4 that the function
sin x
,
x
tends to zero as x → ∞ even though it is oscillating.11 But, generally, some care must
be taken when deciding whether an oscillating function has a limit as the next two
activities illustrate.
11
Of course, in this case it is the fact that the ‘amplitude’ of the oscillations decreases to zero as x → ∞
that guarantees that this limit is zero!
18
2.1. Limits
Activity 2.6
Consider the limits
(i)
lim x(1 + sin x) and (ii)
x→∞
2
lim x(2 + sin x).
x→∞
Do either of these limits exist? If the limit exists, what is it? (Hint: A sketch may
help!)
Activity 2.7
Consider the limits
r
(i)
lim
x→∞
sin x
x
and (ii)
s
sin x .
lim x→∞
x Do either of these limits exist? If the limit exists, what is it? (Hint: A sketch may
help!)
2.1.2
Limits at a point
Limits ‘at a point’ are concerned with the behaviour of a function, f (x), as x tends to
some finite value, say a, a situation we denoted by ‘x → a’ in 174 Calculus. Indeed, as
we saw in that course, this kind of information is useful when we were sketching the
graph of a function because it told us what was happening as x gets ‘very close to a’,
especially if the behaviour of the function as x tends to a from above (denoted by
x → a+ ) was different to its behaviour as x tends to a from below (denoted by x → a− ).
In this section, we consider exactly what this kind of limit means and see how we can
find such a limit (if it exists) in some straightforward cases.
Finite limits
Suppose that l is a real number and that f (x) is a function that is defined for all values
of x < a.12 We start by asking what it means to say that f (x) → l as x → a− , or ‘as x
tends to a from below’, i.e. what it means to say that
lim f (x) = l.
x→a−
Intuitively, we would want to say that this means that we can ensure that f (x) is as
close to l as we want by taking values of x that are less than a but close enough to a.
But, more specifically, we mean that however close we would like f (x) to be to l, there
is some value of x, say X, for which f (x) is as close to l as we wanted if X < x < a.13
12
That is the function needs to be defined for all values of x less than a, but it need not be defined at
x = a.
13
Technically, as you will see in 116 Abstract Mathematics, we say that f (x) → l as x → a− for some
real number, l, if
for any ε > 0, there is a δ > 0 such that, for all x ∈ (a − δ, a), |f (x) − l| < ε.
Here, the value of ε > 0 tells us how close we want to be and, if we can show that there is a δ > 0 such
that |f (x) − l| < ε for all x in the interval (a − δ, a), then we have established that f (x) is as close to
l as we wanted for these values of x below x = a. Indeed, if we can do this for any value of ε, we can
19
2. Limits
Of course, we can also ask what it means to say that f (x) → l as x → a+ , or ‘as x tends
to a from above’, i.e. what it means to say that
2
lim f (x) = l,
x→a+
and, intuitively, we would want to say that this means that we can ensure that f (x) is as
close to l as we want by taking values of x that are greater than a but close enough to a.
But, more specifically, we mean that however close we would like f (x) to be to l, there
is some value of x, say X, for which f (x) is as close to l as we wanted if a < x < X.14
Indeed, if both of the limits
lim f (x)
x→a−
and
lim f (x),
x→a+
exist and, furthermore, they are the same, then we say that the limit
lim f (x),
x→a
exists and is equal to this common value. However, having said that, unless we need to
worry about the limits of f (x) as x → a− and as x → a+ individually,15 we will often be
able to find the limit of f (x) as x → a straightaway. Let’s now consider how we can
actually do this.
Finding finite limits
Again, as we saw with limits at infinity in Section 2.1.1, we can find finite limits by
considering some basic functions that have finite limits and then, by using some
appropriate rules about how limits work, we can then find the limits of certain
combinations of these basic functions. So, we start by stating some finite limits that
arise from basic functions, i.e.
(a) If f (x) = xn where n ∈ N, then f (x) → an as x → a.
(b) If f (x) = cx where c > 0, then f (x) → ca as x → a.
Then we have the rules which tell us how to find the limits of certain combinations of
these functions which are stated in the following theorem.
guarantee that f (x) is getting ‘arbitrarily close’ to l as x → a− . However, we will not make use of this
formal definition here.
14
Technically, as you will see in 116 Abstract Mathematics, we say that f (x) → l as x → a+ for some
real number, l, if
for any ε > 0, there is a δ > 0 such that, for all x ∈ (a, a − δ), |f (x) − l| < ε.
Here, the value of ε > 0 tells us how close we want to be and, if we can show that there is a δ > 0 such
that |f (x) − l| < ε for all x in the interval (a, a − δ), then we have established that f (x) is as close to
l as we wanted for these values of x above x = a. Indeed, if we can do this for any value of ε, we can
guarantee that f (x) is getting ‘arbitrarily close’ to l as x → a+ . However, we will not make use of this
formal definition here.
15
For instance, if it is possible that one of them doesn’t exist or, if both of them exist, it is possible
that they aren’t equal.
20
2.1. Limits
Theorem 2.5 Theorem 2.1 holds if x → ∞ is replaced by x → a− , x → a+ or x → a.
2
Of course, the same caveats apply as the ones we saw after the statement of
Theorem 2.1.
Example 2.10
x2 − 9
.
x→3 x − 3
Find lim (x + 3) and lim
x→3
As we should expect, for the first limit, we have
lim (x + 3) = 3 + 3 = 6,
x→3
as illustrated in Figure 2.4(a) whereas for the second limit we note that, as long as
x 6= 3, we have
(x − 3)(x + 3)
x2 − 9
=
= x + 3,
x−3
x−3
even though this function is not actually defined at x = 3. So, as the limit as x → 3
only considers what the function is doing around x = 3,16 we see that
x2 − 9
= lim (x + 3) = 3 + 3 = 6,
x→3 x − 3
x→3
lim
as illustrated in Figure 2.4(b). In particular, we observe that this limit is 6 even
though this function is undefined at x = 3 and, as such, the point (3, 6) is not part
of the graph of this function.
y
y
y =x+3
6
3
−3
y=
6
x2 −9
x+3
3
O
3
(a)
x
−3
O
3
x
(b)
Figure 2.4: Graphs of the two functions from Example 2.10. (Note that a ‘•’ means that
this point is actually part of the graph of the function whereas a ‘◦’ means that this point
is not actually part of the graph of the function.)
16
That is, whether it is actually defined at x = 3 is irrelevant when we consider the limit as x → 3.
21
2. Limits
Example 2.11
2
x2
.
x→2 x − 3
Find lim
As we anticipate no problems as x → 2, we use Theorem 2.2 to get
lim x2 = 22 = 4 and
x→2
lim (x − 3) = 2 − 3 = −1,
x→2
and, as the second of these limits is non-zero, Theorem 2.2(d) then gives us
x2
4
=
= −4,
x→2 x − 3
−1
lim
as the answer.
Note: As this is fairly obvious, once you have understood the results above, you
would normally just write
x2
22
4
lim
=
=
= −4,
x→2 x − 3
2−3
−1
having taken care to observe that we are not dividing by zero.
Example 2.12
Find lim
x→0
√
1+x−1
.
x
We start by noting that, in this case, we can not simply work out the limit of this
quotient by considering the limits of the numerator and the denominator as x → 0
because both of these limits are zero.
In cases such as this, we employ the useful ‘trick’, called
√ rationalisation, of
multiplying the numerator and the denominator by 1 + x + 1 as this gives us
√
√
√
(1 + x) − 1
1
1+x−1
( 1 + x − 1)( 1 + x + 1)
√
=
= √
=√
.
x
x( 1 + x + 1)
x( 1 + x + 1)
1+x+1
Having done this, we now see that the limits of the numerator and the
as x → 0 are both non-zero, and this gives us
√
1+x−1
1
1
1
lim
= lim √
=√
=
=
x→0
x→0
x
1+1
1+x+1
1+0+1
denominator
1
,
2
as the answer.
Example 2.13
x2 − 3x + 2
Find lim
.
x→1
1 − x2
We start by noting that, in this case, we can not simply work out the limit of this
quotient by considering the limits of the numerator and the denominator as x → 1
because both of these limits are zero.
In cases such as this, where the numerator and the denominator are polynomials, the
22
2.1. Limits
fact that both of them are zero at x = 1 guarantees that they both have x − 1 as a
factor. So, if we employ the useful ‘trick’ of factorising the numerator and
denominator, we see that we have
2
x−2
x − 3x + 2
(x − 1)(x − 2)
=
−
,
=
1 − x2
(1 − x)(1 + x)
1+x
as long as x 6= 1.17 So, as the limit x → 1 only considers what the function is doing
around x = 1, we have
x2 − 3x + 2
1−2
1
x−2
=−
= ,
= − lim
2
x→1
x→1 1 + x
1−x
1+1
2
lim
as the answer.
Infinite limits
If f (x) is a function, we now want to ask what it means to say that this function tends
to infinity as x → a− or as x → a+ , i.e. what it means to say that f (x) → ∞ as x → a−
or as x → a+ which, again abusing our notation, we would write as
lim f (x) = ∞
x→a−
or
lim f (x) = ∞.
x→a+
Of course, intuitively, as we saw in 174 Calculus, we would want to say this means that
the curve y = f (x) has a vertical asymptote given by x = a, i.e. a vertical line that the
curve gets arbitrarily close to. But, more specifically, when we say that
f (x) → ∞ as x → a− we mean that however large we want f (x) to be, let’s say we
want it to be larger than some real number, M , we can find a value of x, say X, for
which f (x) is larger than M if X < x < a.18
f (x) → ∞ as x → a+ we mean that however large we want f (x) to be, let’s say we
want it to be larger than some real number, M , we can find a value of x, say X, for
which f (x) is larger than M if a < x < X.19
17
18
Of course, this function is not actually defined at x = 1.
Technically, as you will see in 116 Abstract Mathematics, we say that f (x) → ∞ as x → a− , if
for any M > 0, there is a δ > 0 such that, for all x ∈ (a − δ, a), f (x) > M .
Here, the value of M > 0 tells us how large we want f (x) to be and, if we can show that there is a δ > 0
such that f (x) > M for all x in the interval (a − δ, a), then we have established that f (x) is always larger
than M for these values of x below x = a. Indeed, if we can do this for any value of M , we can guarantee
that f (x) is getting ‘arbitrarily large’ as x → a− . However, we will not make use of this formal definition
here.
19
Technically, as you will see in 116 Abstract Mathematics, we say that f (x) → ∞ as x → a+ , if
for any M > 0, there is a δ > 0 such that, for all x ∈ (a, a + δ), f (x) > M .
Here, the value of M > 0 tells us how large we want f (x) to be and, if we can show that there is a δ > 0
such that f (x) > M for all x in the interval (a, a + δ), then we have established that f (x) is always larger
than M for these values of x above x = a. Indeed, if we can do this for any value of M , we can guarantee
that f (x) is getting ‘arbitrarily large’ as x → a+ . However, we will not make use of this formal definition
here.
23
2
2. Limits
And, if we find that
lim
2
x→a−
− f (x) = ∞
or
lim
x→a+
− f (x) = ∞,
we say that f (x) tends to minus infinity as x → a− or as x → a+ respectively. That is,
this is what it means to say that f (x) → −∞ as x → a− or as x → a+ which, again
abusing our notation, we could write as
lim f (x) = −∞
x→a−
lim f (x) = −∞,
or
x→a+
respectively.
Finding infinite limits
Let’s start with a simple example.
Example 2.14
Find lim−
x→0
1
1
and lim+ .
x→0 x
x
For the first limit, we see that for values of x < 0, the function 1/x is negative and
so −1/x is positive. Indeed, as these x < 0 get closer to zero, we find that −1/x
tends to infinity, i.e.
1
1
= ∞ =⇒
lim− = −∞.
lim− −
x→0 x
x→0
x
For the second limit, we see that for values of x > 0, the function 1/x is positive
and, as these x > 0 get closer to zero, we find that 1/x tends to infinity, i.e.
lim+
x→0
1
= ∞.
x
Of course, as these two limits are not the same, this means that the limit
1
lim ,
x→0 x
does not exist and, indeed, we can see that this function has a vertical asymptote at
x = 0.
But, generally, as we saw above with finite limits, we can find infinite limits by
considering some basic functions that have infinite limits and then, by using some
appropriate rules about how limits work, we can find the limits of certain combinations
of these basic functions. So, we start by stating some infinite limits that arise from basic
functions, i.e.
(a) If f (x) = xb where b < 0 is a constant, then f (x) → ∞ as x → 0+ .
(b) If f (x) = logd x where d > 1 is a constant, then f (x) → −∞ as x → 0+ .
However, as you will see in Activity 2.8, some care must be taken with the analogue of
(a) when we are considering limits as x → 0− .
24
2.1. Limits
Activity 2.8 Suppose that f (x) = xb where b < 0 is a constant. What can you say
about the limit of this function as x → 0− when (i) b = −1, (ii) b = −2 and (iii)
b = −1/2?
Then, we have the rules which tell us how to find the limits of certain combinations of
these basic functions which are stated in the following theorem.
Theorem 2.6 Theorem 2.3 holds if x → ∞ is replaced by x → a− , x → a+ or x → a.
Of course, the same caveats apply as the ones we saw after the statement of
Theorem 2.3. As an example of why these rules work, we can think of something similar
to Example 2.13, which will no longer give us a finite limit.
Example 2.15
Find lim−
x→1
x2 − 3x + 3
x2 − 3x + 3
and
lim
.
x→1+
1 − x2
1 − x2
We start by noting that, in this case, we can not simply work out the limit of this
quotient by considering the limits of the numerator and the denominator as x → 1−
or as x → 1+ because, in both cases, the limit of the denominator is zero.
In cases such as this, where the denominator is a polynomial, the fact that it is zero
at x = 1 guarantees that it has x − 1 as a factor. So, if we employ the useful ‘trick’
of factorising the denominator, we see that we have
x2 − 3x + 3
x2 − 3x + 3
=
,
1 − x2
(1 − x)(1 + x)
and, of course, this function is not actually defined at x = 1. So, as the limit x → 1−
only considers what the function is doing below x = 1, we have
lim−
x→1
x2 − 3x + 3
x2 − 3x + 3
=
lim
.
x→1− (1 − x)(1 + x)
1 − x2
Now, we can see that, without the 1 − x term in the denominator, we have
x2 − 3x + 3
12 − 3 + 3
1
=
= ,
x→1−
1+x
2
2
lim
and this is positive whereas 1 − x itself tends to zero through positive values as
x → 1− since this gives values of x < 1. Consequently, as we are dividing a positive
number, i.e. 1/2, by smaller and smaller positive numbers, we see that
lim−
x→1
x2 − 3x + 3
x2 − 3x + 3
=
lim
= ∞.
x→1− (1 − x)(1 + x)
1 − x2
Then, applying similar reasoning, we see that as the limit x → 1+ only considers
what the function is doing above x = 1, we have
lim+
x→1
x2 − 3x + 3
x2 − 3x + 3
=
lim
.
x→1+ (1 − x)(1 + x)
1 − x2
25
2
2. Limits
So, we still have
2
x2 − 3x + 3
12 − 3 + 3
1
=
= ,
x→1+
1+x
2
2
and this is positive whereas 1 − x itself now tends to zero through negative values as
x → 1+ since this gives values of x > 1. Consequently, as we are dividing a positive
number, i.e. 1/2, by smaller and smaller [in magnitude] negative numbers, we see
that
x2 − 3x + 3
x2 − 3x + 3
lim+
= −∞.
=
lim
x→1
x→1+ (1 − x)(1 + x)
1 − x2
lim
Of course, as these two limits are not the same, this means that the limit
x2 − 3x + 3
,
x→1
1 − x2
lim
does not exist and, indeed, we can see that this function has a vertical asymptote at
x = 1.
The relationship between infinite and finite limits
We have seen that, as x → a− or as x → a+ , some functions have finite limits and
others have infinite limits but now we want to briefly discuss how these two types of
limit are related. The key result here is the following theorem.
Theorem 2.7 (a) If f (x) → ∞ as x → a− , then
lim−
x→a
1
= 0,
f (x)
and this result also holds if a− is replaced by a+ .
(b) If f (x) → 0 as x → a− and there is a δ ∈ R such that
1
= ∞.
x→∞ f (x)
f (x) > 0 for all a − δ < x < a, then lim
1
= −∞.
x→∞ f (x)
f (x) < 0 for all a − δ < x < a, then lim
The analogue of this result holds if a− is replaced by a+ .
Of course, the results in this theorem should be fairly obvious and we have used them
implicitly in Example 2.15.
Limits that don’t exist
If we have a function f (x) and we find that
lim f (x) = c,
x→a−
where c is a real number or we find that f (x) → ∞ (or −∞) as x → a− , we say that the
limit of f (x) as x → a− exists. And, of course, we can say similar things in the limit as
26
2.1. Limits
x → a+ or x → a. However, not every function has a limit as x → a− or x → a+ or
x → a and, in such cases, we say that this limit does not exist.
Example 2.16
2
Consider the function
(
x+3
if x ≤ 3
f (x) =
.
12 − 3x if x > 3
What can we say about the limit of this function as x → 3?
Here we can see that we may have to worry about what is happening with this
function at x = 3 and so it makes sense to find its limits as x → 3− and as x → 3+ .
In particular, we see that as x → 3− , we are concerned with values of x < 3 that are
getting closer to x = 3 and so we have
lim f (x) = lim− (x + 3) = 3 + 3 = 6,
x→3−
x→3
whereas as x → 3+ , we are concerned with values of x > 3 that are getting closer to
x = 3 and so we have
lim f (x) = lim+ (12 − 3x) = 12 − 3(3) = 3.
x→3+
x→3
So, although both of these limits exist, they are not equal and so the limit of f (x) as
x → 3 does not exist.
Of course, the limits that we have found here are obvious if you look at a sketch of
the graph of this function such as the one in Figure 2.5.
y
y = f (x)
6
3
−3
O
3
4
x
Figure 2.5: The graph of the function f (x) from Example 2.16. (Note that a ‘•’ means
that this point is actually part of the graph of the function whereas a ‘◦’ means that this
point is not actually part of the graph of the function.)
27
2. Limits
2.2
2
Some useful results that involve limits
2.2.1
Continuity
Intuitively, we say that a function is continuous at a point if it has no ‘breaks’ or
‘jumps’ at that point and we can use limits to make this idea more precise. In
particular, if a function, f (x), is such that both
lim f (x)
x→c
and
f (c),
are defined and, furthermore, we have
lim f (x) = f (c),
x→c
then we say that f (x) is continuous at x = c. Indeed, if the function is continuous at
every point in the interval (a, b), we say that it is continuous on that interval and if the
function is continuous at every point in R, we simply say that it is continuous.
Some continuous functions
Most of the basic functions that we considered in 174 Calculus are continuous, in
particular, it should be clear that the following functions are all continuous over their
domains.
Power functions f : R → R where f (x) = xn for n ∈ N.
Exponential functions f : R → (0, ∞) where f (x) = ax for positive a 6= 1.
Trigonometric functions f : R → [−1, 1] where f (x) = sin x or f (x) = cos x.
Logarithmic functions f : (0, ∞) → R where f (x) = loga (x) for positive a 6= 1.
If you are in any doubt about the continuity of any of these functions, you should
consider what we know about their graphs from 174 Calculus.
Some useful results
Having seen which of our basic functions are continuous, we can then see how continuity
is preserved when we take combinations of these functions by using the following
theorem.
Theorem 2.8 If the functions f (x) and g(x) are continuous at x = c, then so are the
functions
kf (x) where k ∈ R;
f (x) + g(x);
f (x)g(x);
f (x)
as long as g(c) 6= 0.
g(x)
28
2.2. Some useful results that involve limits
Moreover, if f (x) is continuous at x = c and g(x) is continuous at d = f (c),20 then the
composition (g ◦ f )(x) = g(f (x)) is also continuous at x = c.
Lastly, suppose that f −1 (x) is the inverse of a function f (x) which is strictly increasing
(or decreasing) over some interval (a, b). If f (x) is continuous at x = c for some
c ∈ (a, b), then f −1 (x) is also continuous at x = f (c).
Let’s take a moment to see how this all works by considering an example.
Example 2.17
Explain why the function
sin x
is continuous for all x 6= 0.
x
We know from above that the functions sin x and x are continuous for all x ∈ R and
so, by Theorem 2.6, we see that the function
sin x
,
x
is also continuous as long as x 6= 0. Indeed, it should be clear that this function is
not continuous at x = 0 since it is not even defined there.
Notice, however, that we can sometimes ‘repair’ failures in continuity by taking a little
more care with the definition of a function as the next example shows.
Example 2.18 Following on from Example 2.17, consider the function f : R → R
given by f (0) = 1 and
sin x
,
f (x) =
x
when x 6= 0. Show that this function is continuous for all x ∈ R.
sin x
[Hint: You may assume that
→ 1 as x → 0. (See Example 2.22.)]
x
We saw in Example 2.17 that f (x) is continuous for all x 6= 0 and, using the hint, we
see that
sin x
lim f (x) = lim
= 1 = f (0),
x→0
x→0 x
which means that f (x) is continuous at x = 0 too.
2.2.2
Differentiability
We saw in 174 Calculus if f (x) is a function, then its derivative, f 0 (x), is the function
defined by
f (x + h) − f (x)
f 0 (x) = lim
,
h→0
h
and, if f 0 (c) exists, we say that f (x) is differentiable at x = c. Indeed, if the function is
differentiable at every point in the interval (a, b), we say that it is differentiable on that
interval and if the function is differentiable at every point in R, we simply say that it is
20
So that, in particular, d is in the range of f which is, in turn, taken to be in the domain of g.
29
2
2. Limits
2
differentiable. Of course, we explored derivatives and what they tell us about functions
in some detail in 174 Calculus and so we will settle for a simple example of how this
works.
Find the derivative of the function f (x) =
Example 2.19
√
x for x > 0.
Using the definition of the derivative, we have
√
√
f (x + h) − f (x)
x+h− x
f (x) = lim
= lim
,
h→0
h→0
h
h
√
where we require that x > 0 so that x is defined. Of course, we can evaluate this
using the rationalisation ‘trick’ that we saw in Example 2.12, i.e. we multiply the
numerator and the denominator of
√
√
x+h− x
,
h
√
√
by x + h + x to get
√
√
√
√ √
√ (x + h) − x
1
x+h− x
x+h− x
x+h+ x
√
=
= √
√
√ =√
√ ,
h
h
x+h+ x
h( x + h + x)
x+h+ x
0
so that we have
√
x+h−
h
√
x
=√
1
1
1
√ = √ ,
√ →√
x+ x
2 x
x+h+ x
as h → 0. This then means that we have
f (x + h) − f (x)
f (x) = lim
= lim
h→0
h→0
h
0
√
x+h−
h
√
x
1
= √ ,
2 x
which is the answer we should have been expecting.
Activity 2.9 Using only the definition, find the derivative of the function
f (x) = x−1 for x 6= 0.
One useful thing to note is that ‘differentiability implies continuity’ in the following
sense.
Theorem 2.9 If the function, f (x), is differentiable at x = c, then it is continuous at
x = c.
To see why this works, consider that if the function, f (x), is differentiable at x = c,
then we know that f 0 (c) exists and is given by
f (c + h) − f (c)
,
h→0
h
which means that if we consider the limit as h → 0 of the function
f 0 (c) = lim
f (c + h) − f (c),
30
2.2. Some useful results that involve limits
we can multiply the numerator and denominator by h to get
f (c + h) − f (c)
= (0)f 0 (c) = 0.
lim f (c + h) − f (c) = lim h
h→0
h→0
h
2
But, this means that
lim f (c + h) = f (c),
h→0
which establishes that f (x) is continuous at x = c.
However, the converse of this theorem is not true as the next activity shows.
Activity 2.10 Show that the function f (x) = |x| is continuous at x = 0, but that it
is not differentiable at that point.
2.2.3
Taylor series and Taylor’s theorem
In 174 Calculus, we saw that the Taylor series for a function, f (x), around some point
x = c was given by
(x − c)2 00
(x − c)n (n)
f (c) + · · · +
f (c) + · · · ,
2!
n!
and, in the case where c = 0, we called this the Maclaurin series for f (x). Indeed, in
that course, we used such series to find approximations to the value of the function f (x)
for values of x close to x = c. In particular, we defined the nth-order approximation to
f (x) around the point x = c to be the polynomial, let’s call it Pn (x), given by
f (x) = f (c) + (x − c)f 0 (c) +
(x − c)n (n)
(x − c)2 00
f (c) + · · · +
f (c).
2!
n!
But, we now want to use a result, called Taylor’s theorem, which will allow us to
investigate how accurate such approximations are.
Pn (x) = f (c) + (x − c)f 0 (c) +
Theorem 2.10 (Taylor’s theorem) If f (x) is a function defined on the interval (a, b)
and all of its derivatives up to f (n+1) (x) exist on (a, b), then for any c, x ∈ (a, b) and any
n ∈ N, we have
(x − c)n+1 (n+1)
f
(d),
f (x) = Pn (x) +
(n + 1)!
for some d lying [strictly] between c and x. In particular, we call
(x − c)n+1 (n+1)
f
(d),
(n + 1)!
the remainder term associated with Pn (x).
Notice that, in Taylor’s theorem, the remainder term tells you about the size of the
difference between f (x) and Pn (x), i.e. the size of the difference between what we are
trying to approximate and the approximation that we have found. As such, if we have
reason to believe that the remainder term is small, then we can be assured that Pn (x) is
a good approximation to f (x). Indeed, using Taylor’s theorem, we can actually find
bounds that determine just how accurate our approximation is as the next example
shows.
31
2. Limits
2
Example 2.20 Find the Taylor series for e−x about x = 0 and use it to find a
third-order approximation to e−1 . Then use Taylor’s theorem to find an upper bound
on the difference between the true value of e−1 and this approximation to it.
If we take f (x) = e−x we have
f 0 (x) = − e−x ,
f 00 (x) = e−x ,
f 000 (x) = − e−x ,
and, spotting the pattern, we see that f (n) (x) = (−1)n e−x for n ≥ 1. Thus, we have
f (0) = 1 and f (n) (0) = (−1)n for n ≥ 1, which means that
−x
e
(x − 0)2
(x − 0)3
(x − 0)n
= 1 + (x − 0)(−1) +
(1) +
(−1) + · · · +
(−1)n + · · ·
2!
3!
n!
n
x
x2 x3
=1−x+
−
+ · · · + (−1)n + · · · ,
2!
3!
n!
is the Taylor series for e−x about x = 0. Indeed, using this, we can see that a
third-order approximation to e−1 is given by
1−1+
1
1 1
1
1
− = − = ,
2! 3!
2 6
3
i.e. our third-order approximation to e−1 is 0.3333 to 4dp.
Referring to Taylor’s theorem, we can then see that the remainder term in this case
is given by
(1 − 0)4 −d e−d
e =
,
4!
24
for some d lying between c = 0 and x = 1. That is, using Taylor’s theorem, we have
found that
1 e−d
e−1 = +
,
3
24
for some d ∈ (0, 1). This means that the difference between the true value of e−1 and
our third-order approximation to it is given by
e−d
1
1
< ,
e−1 − =
3
24
24
as d > 0 means that e−d < 1. Thus, the difference between the true value of e−1 and
our approximation is at most 1/24 (or 0.0417 to 4dp) and so this is the required
upper bound.
Incidentally, the true value of e−1 is 0.3679 to 4dp and so the difference between the
true value of e−1 and our approximation is
1
e−1 − = 0.3679 − 0.3333 = 0.0346,
3
and, as expected, this difference is less than the upper bound that we found for this
quantity.
32
2.2. Some useful results that involve limits
If we extend this idea, we can see that Taylor’s theorem can also be used to find useful
bounds on functions as the next example illustrates.
Example 2.21
2
Use Taylor’s theorem to show that
1 − x < e−x < 1 − x +
x2
,
2
when x > 0.
As we’ll soon see, it makes sense to use the first-order Taylor series for e−x around
x = 0 and the associated remainder term, i.e. using Taylor’s theorem and what we
saw at the beginning of Example 2.20, we have
e−x = 1 − x +
x2 −d
e ,
2
for some d ∈ (0, x) where x > 0. Now, as d > 0, we have e−d < 1 and so we can write
e−x = 1 − x +
x2 −d
x2
e <1−x+ ,
2
2
but, we also know that e−d > 0 for all d ∈ R, and so we can also write
e−x = 1 − x +
x2 −d
e > 1 − x.
2
So, putting these two inequalities together, we get
1 − x < e−x < 1 − x +
x2
,
2
when x > 0, as required.
And, although we won’t dwell too much on it here, we can see that the values of x for
which the Taylor series ‘works’ — an idea we encountered briefly in 174 Calculus — are
those values of x for which the remainder term tends to zero as n → ∞. This is because,
if we have a value of x for which the remainder term tends to zero as n → ∞, then it
should be clear that Pn (x) tends to f (x) as n → ∞. That is, the value of f (x) is the
same as the value of
n
X
(x − c)i (i)
lim Pn (x) = lim
f (c),
n→∞
n→∞
i!
i=0
which is just what we get when we keep all of the terms in the Taylor series.
Using Taylor series to find limits
One particularly important application of Taylor series in this course is that they allow
us to find certain limits. In particular, if we have the Taylor series of some function,
f (x), about the point x = c, we can often use it to deduce the limit of expressions that
involve f (x) as x → c simply because, as x gets closer to c, we would expect f (x) to
33
2. Limits
take values that get closer to the values that arise from its Taylor series. Let’s consider
an example of how this works.
2
Example 2.22
sin x
.
x→0 x
Use the Taylor series for sin x about x = 0 to find lim
We know that the Taylor series for sin x about x = 0 gives us
x3
sin x = x −
+ ··· ,
3!
and so, we have
x3
+ ···
sin x
x2
3!
=
=1−
+ ··· .
x
x
3!
Now, as x → 0, we see that the x2 term (and all the higher-order terms that we have
omitted) should tend to zero,21 leaving us with
sin x
x2
= lim 1 −
+ · · · = 1,
lim
x→0
x→0 x
3!
x−
as the answer.
In particular, we observe that this method is especially useful when we are looking at
limits as x → 0 because we know a lot about the Taylor series of our basic functions
about x = 0 from Section 3.4 of 174 Calculus.
Activity 2.11
Use the appropriate Taylor series to find the following limits.
(a)
1 − cos x
,
x→0
x2
lim
(b)
ln(1 + x)
.
x→0
x
lim
We now consider another important method for working out certain limits and, as we
shall see, this also implicitly relies on the use of Taylor series.
2.2.4
L’Hôpital’s rule
Following on from our earlier discussion of limits, we know that there are still certain
cases which we can not deal with. For instance, suppose that we have to find the limit
lim+
x→a
f (x)
,
g(x)
in the case where f (x) and g(x) both tend to zero as x → a+ . In such cases, we can’t
use Theorem 2.5 because g(x) → 0 as x → a+ but, as long as the stated conditions hold,
we can use L’Hôpital’s rule which runs as follows.
21
If necessary, we can make this idea precise by using the appropriate remainder term, but we will
generally be content with the more ‘intuitive’ calculation that we have presented here.
34
2.2. Some useful results that involve limits
Theorem 2.11 (L’Hôpital’s rule: first form) Suppose that the functions f and g are
differentiable on the interval (a, b) and that g 0 (x) 6= 0 for x ∈ (a, b). If
both lim+ f (x) = 0 and lim+ g(x) = 0, and
x→a
lim+
x→a
x→a
f 0 (x)
= L where L is a real number or ∞ or −∞,22
g 0 (x)
then
lim+
x→a
f (x)
= L,
g(x)
too. Indeed, analogues of this rule hold when lim+ is replaced by
x→a
lim
x→b−
or
lim,
x→c
for some c ∈ (a, b) as well as when we have a is −∞ or b is ∞.
We won’t prove this rule here, but we can see why it works by using Taylor’s theorem.
For instance, looking at f (x) about x = a, Taylor’s theorem dictates that
f (x) = f (a) + (x − a)f 0 (d1 )
for some d1 ∈ (a, x) as we are interested in x ∈ (a, b), whereas looking at g(x) about
x = a, Taylor’s theorem dictates that
g(x) = g(a) + (x − a)g 0 (d2 ),
for some d2 ∈ (a, x) as, again, we are interested in x ∈ (a, b). This means that, looking
at our quotient, we have
f (a) + (x − a)f 0 (d1 )
f (x)
=
.
g(x)
g(a) + (x − a)g 0 (d2 )
But, we are given that
lim f (x) = 0 and
x→a+
lim g(x) = 0,
x→a+
and so we have f (a) = 0 and g(a) = 0,23 which means that our quotient becomes
f (x)
(x − a)f 0 (d1 )
f 0 (d1 )
=
=
,
g(x)
(x − a)g 0 (d2 )
g 0 (d2 )
as we can assume that x 6= a if we are taking the limit as x → a+ . Thus, we have
lim+
x→a
f (x)
f 0 (d1 )
f 0 (x)
= lim+ 0
= lim+ 0
,
g(x) x→a g (d2 ) x→a g (x)
provided that, as assumed in Theorem 2.11, this last limit exists.24
22
That is, this limit exists.
Observe that, using Theorem 2.9, the fact that f and g are differentiable on (a, b) implies that they
must be continuous on (a, b) as well.
24
Observe that the last equality arises because, if d1 and d2 are in the interval (a, x), we can see that
they must both tend to a from above when x → a+ .
23
35
2
2. Limits
2
Activity 2.12 Given that f (x) and g(x) are differentiable functions which both
tend to zero as x → ∞, use the substitution x = 1/t and Theorem 2.11 to show that
f 0 (x)
f (x)
= lim 0
,
x→∞ g (x)
x→∞ g(x)
lim
provided that the limit on the right-hand side exists.
Let’s now look at some examples of how L’Hôpital’s rule can be applied.
Example 2.23
1 − cos x
ln(1 + x)
and lim
.
x→0
x→0
x
x
Use L’Hôpital’s rule to find lim
For the first limit, we note that the numerator and the denominator both tend to
zero as x → 0 and we also have
f (x)
1 − cos x
=
g(x)
x
=⇒
f 0 (x)
sin x
=
,
0
g (x)
1
where, as x → 0, the limit of this second quotient is zero. So, using L’Hôpital’s rule,
we have
sin x
1 − cos x
= lim
= 0,
lim
x→0 1
x→0
x
as the answer.
For the second limit, we again note that the numerator and the denominator both
tend to zero as x → 0 and we also have
f (x)
ln(1 + x)
=
g(x)
x
=⇒
f 0 (x)
1/(1 + x)
=
,
0
g (x)
1
where, as x → 0, the limit of this second quotient is one. So, using L’Hôpital’s rule,
we have
ln(1 + x)
1/(1 + x)
lim
= lim
= 1,
x→0
x→0
x
1
as the answer in agreement with what we saw in Activity 2.11(b)
Activity 2.13
Following on from Example 2.22, use L’Hôpital’s rule to verify that
sin x
= 1.
x→0 x
lim
Another case where we run into problems involves the limit
lim+
x→a
f (x)
,
g(x)
in the case where f (x) and g(x) both tend to infinity as x → a+ . In such cases, we can’t
use Theorem 2.7 because this case isn’t covered but, as long as the stated conditions
hold, we can use another form of L’Hôpital’s rule which runs as follows.
36
2.2. Some useful results that involve limits
Theorem 2.12 (L’Hôpital’s rule: second form) Suppose that the functions f and g
are differentiable on the interval (a, b) and that g 0 (x) 6= 0 for x ∈ (a, b). If
both lim+ f (x) = ∞ and lim+ g(x) = ∞, and
x→a
lim+
x→a
x→a
f 0 (x)
= L where L is a real number or ∞ or −∞,25
g 0 (x)
then
f (x)
= L,
x→a g(x)
too. Indeed, analogues of this rule hold when lim+ is replaced by
lim+
x→a
lim
x→b−
or
lim,
x→c
where c ∈ (a, b) as well as when we have a is −∞ or b is ∞.
To get a sense of why this works observe that, if the limits of f (x) and g(x) are both
infinity as x → a+ , we can see that 1/f (x) and 1/g(x) both tend to zero as x → a+ and
so, using Theorem 2.11, we have
2 f (x)
1/g(x)
g 0 (x)/[g(x)]2
f (x)
g 0 (x)
lim
= lim+
= lim+ 0
= lim+
lim
,
x→a+ g(x)
x→a 1/f (x)
x→a f (x)/[f (x)]2
x→a g(x)
x→a+ f 0 (x)
assuming that these limits exist. Then, if we also have (for the sake of argument) some
L 6= 0 such that
f 0 (x)
lim+ 0
= L,
x→a g (x)
we see that this gives us
f (x)
= L,
lim+
x→a g(x)
too. Of course, this isn’t a proof of Theorem 2.12, but it does indicate how it is related
to Theorem 2.11 if we aren’t being that careful. Let’s look at an example of how it can
be applied.
Example 2.24
Use L’Hôpital’s rule to find lim
x→∞
ln(1 + x)
.
x
We note that the numerator and the denominator both tend to infinity as x → ∞
and we also have
ln(1 + x)
f (x)
=
g(x)
x
=⇒
f 0 (x)
1/(1 + x)
=
,
0
g (x)
1
where, as x → ∞, the limit of this second quotient is zero. So, using L’Hôpital’s rule,
we have
ln(1 + x)
1/(1 + x)
lim
= lim
= 0,
x→∞
x→∞
x
1
as the answer.
25
That is, this limit exists.
37
2
2. Limits
Activity 2.14
Use L’Hôpital’s rule to find lim x
x→∞
2
π
2
− tan−1 x .
Lastly, as you will see in the next activity, some care must be taken when finding limits
with L’Hôpital’s rule as it is not always applicable.
Activity 2.15
Find the limits
(a)
lim+
x→0
ln x
x
and (b)
x + sin x
.
x→∞
x
lim
Why is L’Hôpital’s rule not applicable for either of these limits?
Learning outcomes
At the end of this chapter and having completed the relevant reading and activities, you
should be able to:
find limits or explain why they don’t exist;
assess whether a function is continuous;
find derivatives using the definition of the derivative;
use Taylor series to find limits and Taylor’s theorem to find bounds;
use L’Hôpital’s rule to find limits.
Solutions to activities
Solution to activity 2.1
We are given that f (x) → l and g(x) → m as x → ∞ where l, m ∈ R, and so using
Theorem 2.1(a) with c = −1 and (b), we see that
f (x) − g(x) = f (x) + [−g(x)] → l + [−m] = l − m as x → ∞,
and, for any constants c and d, we also have
cf (x) + dg(x) = [cf (x)] + [dg(x)] → cl + dm as x → ∞,
again using Theorem 2.1(a) and (b).
Solution to activity 2.2
For (a), we follow the method in Example 2.3 and divide the numerator and
denominator by the highest power of x that occurs in the given quotient. In this case,
the highest power of x in the quotient is x3 and so we get
x2 + x + 1
(1/x) + (1/x2 ) + (1/x3 )
0+0+0
0
=
lim
=
= = 0.
2
x→∞
x→∞ x + x2 + x3
(1/x ) + (1/x) + 1
0+0+1
1
lim
38
2.2. Solutions to activities
For (b), we note that the highest power of x in the quotient is x1/2 and so, using the
same method as in (a), we get
2
1/2
1 + sinx x − x2
(x + sin x − 2)1/2
.
lim √
= lim
x→∞
x→∞ 1 + sin
√ x − √2
x + sin x − 2
x
x
Now, from Example 2.4, we know that
sin x
= 0,
x→∞ x
lim
and so, as x → ∞, the numerator has a limit given by (1 + 0 + 0)1/2 = 1 if we use
Theorem 2.1(e). Further, in the denominator, we see that
sin x
lim √ = 0,
x→∞
x
if, following Example 2.4, we use the Sandwich theorem with the inequality
1
sin x
1
−√ ≤ √ ≤ √ ,
x
x
x
which clearly holds for x > 0. So, overall, we have
1/2
1 + sinx x − x2
(x + sin x − 2)1/2
1
lim √
= lim
=
= 1,
sin
x
2
x→∞
x→∞ 1 + √ − √
1+0+0
x + sin x − 2
x
x
as our final answer.
Solution to activity 2.3
Given that f (x) → ∞ as x → ∞, we see that the analogue of Theorem 2.3(a) when
c < 0 is
cf (x) → −∞ as x → ∞,
whereas, given that g(x) → m as x → ∞ where m < 0 is a real number, we see that the
analogues of Theorem 2.3(d)-(f) are
f (x) + g(x) → ∞,
f (x)g(x) → −∞ and
f (x)
→ −∞,
g(x)
as x → ∞.
We also see that, if 0 < d < 1, we have (say) ln d < 0 and ln x → ∞ as x → ∞ so, using
the ‘change of base formula’ for logarithms, we have
logd x =
ln x
→ −∞,
ln d
as x → ∞ using our analogue of Theorem 2.3(a) from above.
39
2. Limits
Solution to activity 2.4
2
For (i), we consider the analogue of Theorem 2.3(a) and note that regardless of the fact
that f (x) → ∞ as x → ∞, if we take c = 0, we have
cf (x) = 0 → 0 as x → ∞.
For (ii), we see that if f (x) → ∞ and g(x) → −∞ as x → ∞, then the analogue of
Theorem 2.3(b) tells us nothing since we can make no sense of ‘∞ − ∞’. But, the
analogue of Theorem 2.3(c) is
f (x)g(x) → −∞ as x → ∞,
as the result of the product will be negative and arbitrarily large in magnitude.
For (iii), we see that if f (x) → ∞ and g(x) → 0 as x → ∞, then the analogue of
Theorem 2.3(d) is
f (x) + g(x) → ∞ as x → ∞,
but the analogue of Theorem 2.3(e) tells us nothing since we can make no sense of
‘∞ · 0’. The analogue of Theorem 2.3(f) also tells us nothing in this case as we can make
no sense of ‘∞/0’.26
Solution to activity 2.5
For (a), notice that, as it stands, we can make no sense of x2 − x3 as x → ∞ because we
can make no sense of ‘∞ − ∞’. However, if we consider that the highest power of x here
is x3 (i.e. this is the ‘dominant term’), we see that using the analogue of Theorem 2.3(e)
that we saw in Activity 2.3, we have
1
2
3
3
− 1 = −∞,
lim (x − x ) = lim x
x→∞
x→∞
x
since, as x → ∞, the first term in the product tends to ∞ whereas the second term
term tends to −1.
For (b), following on from what we saw in Example 2.6, we see that dividing the
numerator and denominator by the highest power of x in the denominator, i.e. x, we get
sin x
lim x −
x→∞
x − sinx x
x2 − sin x
x
=
lim
= lim
= ∞,
x→∞ x + sin x
x→∞ 1 + sin x
1
x
if we use Theorem 2.3(f) and the result from Example 2.4.
26
However, as we will see in Theorem 2.4, we can make some sense of this if we have some information
about the sign of g(x).
40
2.2. Solutions to activities
Solution to activity 2.6
For (i), let’s consider the function
2
f (x) = x(1 + sin x),
for x ≥ 0 so that, since −1 ≤ sin x ≤ 1, we have
0 ≤ 1 + sin x ≤ 2
0 ≤ x(1 + sin x) ≤ 2x
=⇒
=⇒
0 ≤ f (x) ≤ 2x,
and so the graph of the curve y = f (x) is always between the straight lines y = 0 (i.e.
the x-axis) and y = 2x. In particular, given the sin x term in f (x), we would expect this
curve to ‘oscillate’ between these two straight lines. Consequently, we should anticipate
that a rough sketch of this curve would look something like the one in Figure 2.6(a).
Indeed, with this sketch, it should be clear that the limit
lim x(1 + sin x),
x→∞
does not exist since, as x → ∞, f (x) neither stays arbitrarily close to any finite value
(i.e. there is clearly no finite limit) nor is it always getting arbitrarily large (i.e. it
clearly doesn’t tend to infinity).
For (ii), let’s consider the function
g(x) = x(2 + sin x),
for x ≥ 0 so that, since −1 ≤ sin x ≤ 1, we have
1 ≤ 2 + sin x ≤ 3
=⇒
x ≤ x(2 + sin x) ≤ 3x
=⇒
x ≤ g(x) ≤ 3x,
and so the graph of the curve y = g(x) is always between the straight lines y = x and
y = 3x. In particular, given the sin x term in g(x), we would expect this curve to
‘oscillate’ between these two straight lines. Consequently, we should anticipate that a
rough sketch of this curve would look something like the one in Figure 2.6(b). Indeed,
with this sketch, it should be clear that we have
lim x(2 + sin x) = ∞,
x→∞
as, for x ≥ 0, g(x) ≥ x and so g(x) must tend to infinity as x → ∞.
Solution to activity 2.7
For (i), we see that the limit
r
lim
x→∞
sin x
,
x
does not exist because, although
sin x
= 0,
x→∞ x
lim
as we saw in Example 2.4, we cannot take the limit of
r
sin x
,
x
41
2. Limits
2
(a)
(b)
Figure 2.6: The sketches for Activity 2.6. (a) The dashed line is y = 2x and the solid
curve that ‘oscillates’ between this line and the x-axis (i.e. the line y = 0) is y = f (x).
(b) The dotted line is y = x, the dashed line is y = 3x and the solid curve that ‘oscillates’
between these two lines is y = g(x).
as x → ∞ because this is not a well-defined function from R to R. In particular, we see
that if we want to consider its values (and whether they are staying arbitrarily close to
some limiting number — like, say, zero) for suitably large values of x, we find that for
some of those values of x, sin x will be negative and so the square root will not return a
value (i.e. a real number) for comparison.27 See, for instance, the rough sketch in
Figure 2.7(a).
For (ii), we see that the function
s
sin x ,
f (x) = x is well-defined for all x > 0 and so, unlike with (i), we can begin to make sense of the
limit of f (x) as x → ∞. Indeed, since 0 ≤ | sin x| ≤ 1 for all x > 0, we can see that
1
0 ≤ f (x) ≤ √ ,
x
√
and so, as 1/ x → 0 as x → ∞, we can conclude
s
sin x = 0,
lim x→∞
x 27
Notice, in particular, that this is a case where Theorem 2.1(e) fails because with b = 1/2 and
l = lim f (x) = lim
x→∞
x→∞
sin x
= 0,
x
we can not make sense of
lim [f (x)]b ,
x→∞
even though lb = 01/2 = 0. This is why we insist that l > 0 in the statement of the theorem. (Because
l > 0 guarantees that f (x) > 0 for suitably large values of x and so we can meaningfully find the
corresponding values of, say, [f (x)]1/2 .)
42
2.2. Solutions to activities
using Theorem 2.2.28 Of course, this can be seen very easily if we consider the rough
sketch in Figure 2.7(b).
2
(a)
(b)
√
Figure 2.7: The sketches for Activity 2.7. (a) The dashed curve is y = 1/ x and the
solid
p curve that ‘oscillates’ between this curve and the x-axis (i.e. the line y = 0) is y =
(sin x)/x where it exists! In particular, for (2k −1)π < x < 2kπ with k ∈ N, although
it may look like this curve is giving us y = 0 in the sketch, this is misleading because
it
√
does not actually exist for these values of x! (b) The dashed curve is y = 1/ x and
the solid curve that ‘oscillates’ between this curve and the x-axis (i.e. the line y = 0) is
y = f (x).
Solution to activity 2.8
For (i), we have f (x) = x−1 , and as we saw in Example 2.14, we have f (x) → −∞ as
x → 0− .
For (ii), we have f (x) = x−2 , and thinking about this in light of Example 2.14, we see
that when x < 0, the function 1/x2 is positive. Indeed, as we take values of x < 0 that
get closer to zero, we find that 1/x2 takes larger and larger positive values. That is,
f (x) → ∞ as x → 0− in this case.
For (iii), we have f (x) = x−1/2 and, if x < 0, this is not a well-defined function from R
to R as we have to take the square root of a negative number. Consequently, the limit of
f (x) as x → 0− can not exist either.
Solution to activity 2.9
Given the function f (x) = x−1 for x 6= 0, we use the definition of the derivative to see
that
(x + h)−1 − x−1
f (x + h) − f (x)
= lim
,
f 0 (x) = lim
h→0
h→0
h
h
where we have x 6= 0 so that x−1 is defined. Of course, we can then see that
1
1
−
(x + h)−1 − x−1
−h
−1
= x+h x =
=
,
h
h
x(x + h)h
x(x + h)
28
In particular, with reference to the previous footnote, we still can’t use Theorem 2.1(e) here because
‘l’ is still zero!
43
2. Limits
2
as we have h 6= 0 when we are considering the limit as h → 0. This then means that we
have
1
−1
= − 2,
f 0 (x) = lim
h→0 x(x + h)
x
which is the answer we should have been expecting.
Solution to activity 2.10
We see that the function f (x) = |x| is continuous at x = 0 because
lim f (x) = lim |x| = |0| = 0 = f (0).
x→0
x→0
However, if we consider the definition of the derivative at x = 0, we need to look at
|h| − 0
|h|
f (0 + h) − f (0)
=
=
,
h
h
h
so that, when h < 0, we have |h| = −h and
lim−
h→0
f (0 + h) − f (0)
|h|
−h
= lim−
= lim−
= lim− (−1) = −1,
h→0
h→0
h→0
h
h
h
whereas, when h > 0, we have |h| = h and
lim+
h→0
f (0 + h) − f (0)
|h|
h
= lim+
= lim+ = lim+ 1 = 1.
h→0
h→0 h
h→0
h
h
Consequently, we see that f (x) is not differentiable at x = 0 because
f (0 + h) − f (0)
,
h→0
h
f 0 (0) = lim
does not exist since the limits as h → 0− and as h → 0+ are not equal.
Solution to activity 2.11
For (a), we know that the Taylor series for cos x about x = 0 gives us
cos x = 1 −
x2 x4
+
− ··· ,
2!
4!
and so, we have
x2 x 4
1− 1−
+
− ···
1 − cos x
2!
4!
=
x2
x2
=
1
x2
−
+ ··· .
2!
4!
Now, as x → 0, we see that the x2 term (and all the higher-order terms that we have
omitted) should tend to zero, leaving us with
1 − cos x
1
x2
1
lim
= lim
−
+ ··· = ,
2
x→0
x→0
x
2!
4!
2
as the answer.
44
2.2. Solutions to activities
For (b), we know that the Taylor series for ln(1 + x) about x = 0 gives us
ln(1 + x) = x −
x2
+ ··· ,
2
2
and so, we have
x2
+ ···
ln(1 + x)
x
2
=
= 1 − + ··· .
x
x
2
Now, as x → 0, we see that the x term (and all the higher-order terms that we have
omitted) should tend to zero, leaving us with
x−
ln(1 + x)
x
lim
= lim 1 − + · · · = 1,
x→0
x→0
x
2
as the answer.
Solution to activity 2.12
Essentially, the substitution x = 1/t gives us t → 0+ as x → ∞ and
f (x)
f (1/t)
= lim+
,
x→∞ g(x)
t→0 g(1/t)
lim
where, as f (x) and g(x) both tend to zero as x → ∞, we see that the functions f (1/t)
and g(1/t) must also both tend to zero as t → 0+ . So, using the chain rule,
Theorem 2.11 gives us
lim+
t→0
f 0 (x)
f 0 (1/t)(−1/t2 )
f 0 (1/t)
f (1/t)
= lim+ 0
=
lim
=
lim
,
g(1/t) t→0 g (1/t)(−1/t2 ) t→0+ g 0 (1/t) x→∞ g 0 (x)
if we use the substitution x = 1/t again. So, putting this all together, we have shown
that
f (x)
f 0 (x)
lim
= lim 0
,
x→∞ g(x)
x→∞ g (x)
provided that the limit on the right-hand side exists.
Solution to activity 2.13
To find the given limit, we note that the numerator and the denominator both tend to
zero as x → 0 and we also have
f (x)
sin x
=
g(x)
x
=⇒
f 0 (x)
cos x
=
,
g 0 (x)
1
where, as x → 0, the limit of this second quotient is one. So, using L’Hôpital’s rule, we
have
sin x
cos x
lim
= lim
= 1,
x→0 x
x→0
1
as the answer in agreement with what we saw in Example 2.22.
45
2. Limits
Solution to activity 2.14
2
In order to use L’Hôpital’s rule to find the given limit, we first need to write the given
product as a quotient. Of course, the sensible way to do this is to write it as
lim x
x→∞
π
2
− tan−1 x = lim
x→∞
π
2
− tan−1 x
,
1/x
so that we now have a quotient where the numerator and the denominator both tend to
zero as x → ∞. We also have
f (x)
=
g(x)
π
2
− tan−1 x
1/x
=⇒
f 0 (x)
−1/(1 + x2 )
x2
=
=
=
g 0 (x)
−1/x2
1 + x2
1
x2
1
,
+1
where, as x → ∞, the limit of this second quotient is one. So, using L’Hôpital’s rule, we
have
π
π
− tan−1 x
−1/(1 + x2 )
x2
−1
2
lim x
− tan x = lim
= lim
=
lim
= 1,
x→∞
x→∞
x→∞
x→∞ 1 + x2
2
1/x
−1/x2
as the answer.
Solution to activity 2.15
For (a), L’Hôpital’s rule is not applicable because, even though we have a quotient, the
numerator tends to minus infinity as x → 0+ whereas the denominator tends to zero as
x → 0+ . However, we can see that
1
ln x
= lim+
ln x = −∞,
lim+
x→0
x→0
x
x
using the analogue of Theorem 2.3(c) that we saw in Activity 2.4(ii).
For (b), we see that L’Hôpital’s rule is not applicable because, even though we have a
quotient where the numerator and denominator both tend to infinity as x → ∞, we find
that
f (x)
x + sin x
f 0 (x)
1 + cos x
=
=⇒
=
,
0
g(x)
x
g (x)
1
and, as x → ∞, the limit of this second quotient does not exist. However, we can see
that
x + sin x
sin x
lim
= lim 1 +
= 1 + 0 = 1,
x→∞
x→∞
x
x
if we use the result from Example 2.4.
Exercises
Exercise 2.1
Find functions f and g which are such that, as x → ∞, f (x) → ∞ and g(x) → 0 but
(a)
46
f (x)g(x) → ∞,
2.2. Solutions to exercises
(b)
f (x)g(x) → 0,
(c)
f (x)g(x) → −2,
(d)
the limit of f (x)g(x) does not exist.
2
Exercise 2.2
Evaluate the following limits.
2
1/2
(a) lim (x + x) − x ,
x→∞
(b)
3
2 1/3
lim (x + x )
x→∞
−x .
Exercise 2.3
Use L’Hôpital’s rule to determine the following limits.
(a)
ln(1 − cos x)
,
x→π/2
cos x
lim
(b)
lim (1 − cos x)tan x .
x→π/2
Exercise 2.4
Using only the definition, find the derivative of the function f (x) = ln x for x > 0.
Exercise 2.5
For x 6= 0, consider the function
f (x) =
1 − cos(2x)
.
2x2
Find the limit of f (x) as x → 0 and hence define a function g(x) that is continuous for
all x ∈ R which is the same as f (x) for all x 6= 0. Hence use the definition of the
derivative to find g 0 (0).
Solutions to exercises
Solution to exercise 2.1
We can easily see that if we pick the functions f and g as follows they will satisfy the
requirement that, as x → ∞, f (x) → ∞ and g(x) → 0, as well as the additional
requirement stated in the question.
1
gives us f (x)g(x) = x → ∞ as x → ∞.
x
(a)
f (x) = x2 and g(x) =
(b)
f (x) = x and g(x) =
(c)
f (x) = x and g(x) = −
1
1
gives
us
f
(x)g(x)
=
→ 0 as x → ∞.
x2
x
2
gives us f (x)g(x) = −2 → −2 as x → ∞.
x
47
2. Limits
sin x
gives us f (x)g(x) = sin x.
x
Of course, in (d), the limit of f (x)g(x) does not exist as x → ∞.
(d)
2
f (x) = x and g(x) =
The lesson here is that, as we observed in Activity 2.4(iii), if we only know that
f (x) → ∞ and g(x) → 0 as x → ∞, we can infer nothing about the limit of f (x)g(x) as
x → ∞.
Solution to exercise 2.2
For (a), we can use the rationalisation ‘trick’ from Example 2.12 to see that
(x2 + x)1/2 − x =
[(x2 + x)1/2 − x][(x2 + x)1/2 + x]
(x2 + x) − x2
x
=
= 2
.
2
1/2
2
1/2
(x + x) + x
(x + x) + x
(x + x)1/2 + x
We then observe that, as in Example 2.3, we can divide the numerator and the
denominator by the highest power of x that occurs in the quotient, i.e. x itself, to get
(x2
x
1
1
1
=
→
= ,
1/2
1/2
+ x) + x
1+1
2
1 + x1
+1
as x → ∞. That is, we have shown that
1
2
1/2
lim (x + x) − x = .
x→∞
2
For (b), we can also rationalise29 to get
(x3 + x2 )1/3 − x =
[(x3 + x2 )1/3 − x][(x3 + x2 )2/3 + x(x3 + x2 )1/3 + x2 ]
(x3 + x2 )2/3 + x(x3 + x2 )1/3 + x2
=
(x3 + x2 ) − x3
(x3 + x2 )2/3 + x(x3 + x2 )1/3 + x2
=
x2
.
(x3 + x2 )2/3 + x(x3 + x2 )1/3 + x2
We can then divide the numerator and the denominator by the highest power of x that
occurs in the quotient, i.e. x2 , to get
x2
1
1
1
→
=
= ,
2/3
1/3
3
2
2/3
3
2
1/3
2
(x + x ) + x(x + x ) + x
1+1+1
3
+ 1 + x1
+1
1 + x1
29
The rationalisation ‘trick’ in (a) takes ‘a − b’ and uses the well-known difference of two squares, i.e.
(a − b)(a + b) = a2 − b2 ,
to ‘remove’ the square root. In (b), the rationalisation ‘trick’ also takes ‘a − b’ but we need to use the
less well-known difference of two cubes, i.e.
(a − b)(a2 + ab + b2 ) = a3 − b3 ,
to ‘remove’ the cube root.
48
2.2. Solutions to exercises
as x → ∞. That is, we have shown that
1
3
2 1/3
lim (x + x ) − x = .
x→∞
3
2
Incidentally, notice that, if we take n ∈ N we can consider the function
"
#
1/n
1
(xn + xn−1 )1/n − x = x 1 +
−1 ,
x
which replicates what we had in (a) and (b) if we take n = 2 and n = 3 respectively.
Now, for suitably large x, 1/x is close to zero and so we can use the binomial theorem30
to see that
1/n
1 1
( − 1)
1
1
n n
1+
=1+
+
+ ··· ,
x
nx
2! x2
which means that
n
(x + x
n−1 1/n
)
−x=x
1 1
1 1
( − 1)
( − 1)
1
1
n n
n n
+
+
·
·
·
−
1
=
+
+ ··· .
1+
2
nx
2! x
n
2! x
Consequently, as x → ∞, we find that the term in 1/x (and all of the neglected terms)
tend to zero, leaving us with
1
2
1/n
lim (x + x) − x = ,
x→∞
n
in agreement with what we saw above.
Solution to exercise 2.3
For (a), we see that the numerator and the denominator both tend to zero as x → π/2
and we also have
ln(1 − cos x)
f (x)
=
g(x)
cos x
=⇒
f 0 (x)
(sin x)/(1 − cos x)
−1
=
=
,
g 0 (x)
− sin x
1 − cos x
where, as x → π/2, the limit of this second quotient is −1. So, using L’Hôpital’s rule,
we have
ln(1 − cos x)
(sin x)/(1 − cos x)
−1
= lim
= lim
= −1,
x→π/2
x→π/2
x→π/2 1 − cos x
cos x
− sin x
lim
as the answer.
For (b), we note that we can write
(1 − cos x)tan x = e(tan x) ln(1−cos x) ,
and, as x → π/2, we have
(tan x) ln(1 − cos x) = (sin x)
30
ln(1 − cos x)
→ (1)(−1) = −1,
cos x
See Example 3.25 in Section 3.4.1 of 174 Calculus.
49
2. Limits
if we use the answer to (a). Consequently, we find that
lim (1 − cos x)tan x = lim e(tan x) ln(1−cos x) = e−1 ,
2
x→π/2
x→π/2
is the answer.
Solution to exercise 2.4
We are given f (x) = ln x and, for x > 0, the definition of the derivative gives us
f (x + h) − f (x)
ln(x + h) − ln x
= lim
,
h→0
h→0
h
h
f 0 (x) = lim
and, as the numerator and the denominator of this quotient both tend to zero as h → 0,
we can use L’Hôpital’s rule to see that
1
1
1
= ,
f 0 (x) = lim x + h = lim
h→0
h→0
1
x+h
x
which is the answer we should expect.
Solution to exercise 2.5
We are given the function
1 − cos(2x)
,
2x2
for x 6= 0 and so we can see that, as the numerator and the denominator of this quotient
tend to zero as x → 0, we can use L’Hôpital’s rule to see that
f (x) =
lim f (x) = lim
x→0
x→0
1 − cos(2x)
2 sin(2x)
sin u
= lim
= lim
= 1,
2
x→0
u→0
2x
4x
u
if we let u = 2x and use the result from Example 2.22. This means that if we define the
function, g : R → R, to be such that g(0) = 1 and g(x) = f (x) for x 6= 0, we will have a
function that is continuous for all x ∈ R. In particular, g(x) is continuous for all x 6= 0
because f (x) is by Theorem 2.8 and, as
lim g(x) = lim f (x) = 1 = g(0),
x→0
x→0
g(x) is continuous at x = 0 too.
We can now use the definition of the derivative to see that
g(0 + h) − g(0)
= lim
g 0 (0) = lim
h→0
h→0
h
1 − cos(2h)
−1
1 − cos(2h) − 2h2
2h2
= lim
,
h→0
h
2h3
and, as the numerator and the denominator of this quotient tend to zero as h → 0, we
can use L’Hôpital’s rule to see that
0 + 2 sin(2h) − 4h
sin(2h) − 2h
= lim
,
2
h→0
h→0
6h
3h2
g 0 (0) = lim
50
2.2. Solutions to exercises
and, as the numerator and the denominator of this quotient also tend to zero as h → 0,
we can use L’Hôpital’s rule again to see that
2
cos(2h) − 1
2 cos(2h) − 2
= lim
,
h→0
h→0
6h
3h
g 0 (0) = lim
and, as the numerator and the denominator of this quotient also tend to zero as h → 0,
we can use L’Hôpital’s rule yet again to see that
−2 sin(2h)
= 0,
h→0
3
g 0 (0) = lim
is the final answer. Of course, this is obvious if we use the Taylor series for cos(2h)
about h = 0 to observe that
(2h)2 (2h)4
2
2
1 − cos(2h) − 2h = 1 − 1 −
+
+ · · · − 2h2 = − h4 + · · · ,
2!
4!
3
so that we get
4
− 2h3 + · · ·
1 − cos(2h) − 2h2
1
g (0) = lim
=
lim
= lim − h = 0,
3
3
h→0
h→0
h→0
2h
2h
3
0
as the final answer.
51
2. Limits
2
52
Chapter 3
The Riemann integral
3
Essential reading
(For full publication details, see Chapter 1.)
+ Binmore and Davies (2002) Sections 10.2–10.4.
+ Ostaszewski (1991) Sections 17.1–17.3.
Further reading
+ Adams and Essex (2010) pp.90–95 and 102–104.
+ Wrede and Spiegel (2010) Sections 5.2–5.5.
Aims and objectives
The objectives of this chapter are:
to introduce the Riemann integral and see how it allows us to define a definite
integral in terms of the area it represents;
to see how the Riemann integral and differentiation are related via the
Fundamental Theorem of Calculus.
Specific learning outcomes can be found near the end of this chapter.
3.1
The Riemann integral
In Section 5.3 of 174 Calculus we interpreted the definite integral
Z b
f (x) dx,
a
where the integrand, f (x), is continuous and non-negative over the interval [a, b] as the
area of the region between the curve y = f (x), the x-axis and the vertical lines x = a
and x = b. Indeed, we saw how to find such integrals, and hence the corresponding
areas, by using the idea that integrals can be seen as antiderivatives. However, some
definite integrals, such as
Z
1
2
e−x dx,
0
53
3. The Riemann integral
2
can not be found in this way as the integrand, which is e−x in this case, has no
antiderivative1 and so, in such cases, we would need another way of interpreting what
this integral is. One way of doing this would be to consider the area of the region
2
between the curve y = e−x , the x-axis and the vertical lines x = 0 and x = 1 directly,
and having done this, we could interpret this area as the value of this integral.
3
Indeed, in this chapter, we will see how to define the definite integral
Z
b
f (x) dx,
a
in terms of the area of the region between the curve y = f (x), the x-axis and the
vertical lines x = a and x = b and we will explore some properties of this integral. In
particular, we will call this kind of integral a Riemann integral.
3.1.1
Lower and upper estimates of an area
Suppose that we have a non-negative continuous function, f (x), over some finite interval
[a, b] and we want to estimate the area, A, of the region between the curve y = f (x), the
x-axis and the vertical lines x = a and x = b as illustrated in Figure 3.1(a). One way to
do this is to take a partition, P, of the interval [a, b], i.e. a finite set of points
P = {x0 , x1 , x2 , . . . , xn−1 , xn } where a = x0 < x1 < x2 < · · · < xn = b,
and use this to divide the interval [a, b] into n sub-intervals of the form [xi−1 , xi ] for
i = 1, . . . , n as illustrated in Figure 3.1(b).
y
y
y = f (x)
y = f (x)
A
O a
...
b
(a)
x
O x
0
x1
x2
...
xn−1 xn
x
(b)
Figure 3.1: (a) The area, A, of the region between the curve y = f (x), the x-axis and the
vertical lines x = a and x = b. (b) The partition, P, of the interval [a, b] where x0 = a
and xn = b. Note that, for clarity, only the first three and last two points of the partition
are shown.
1
By this we mean that there is no combination of our basic functions which can be differentiated to
2
give the function e−x .
54
3.1. The Riemann integral
Each of these sub-intervals can then be taken to be the base of a rectangle whose height
in some way depends on the values of f (x) in that sub-interval. In particular, we can
take the height of the rectangle in each sub-interval to be the
minimum value of f (x) in that sub-interval so that the sum of the areas of these
rectangles gives us a lower estimate of the area, or the
maximum value of f (x) in that sub-interval so that the sum of the areas of these
rectangles gives us an upper estimate of the area.
Clearly, these rectangles will allow us to estimate the area of the region involved.
Specifically, we can see that the
lower estimate, L(P), will be the sum of the areas of the rectangles illustrated in
Figure 3.2(a), i.e.
n
X
L(P) =
Li ,
i=1
and this will clearly be less than the area, A, that we are after as the rectangles
give us ‘too little’ area if we compare their area with the area in Figure 3.1(a).
upper estimate, U (P), will be the sum of the areas of the rectangles illustrated in
Figure 3.2(b), i.e.
n
X
U (P) =
Ui ,
i=1
and this will clearly be greater than the area, A, that we are after as the rectangles
give us ‘too much’ area if we compare their area with the area in Figure 3.1(a).
y
y
y = f (x)
L1
O x
0
L2
x1
x2
...
Ln
...
xn−1 xn
(a)
y = f (x)
U1
x
O x
0
U2
x1
x2
...
Un
...
xn−1 xn
x
(b)
Figure 3.2: In (a) and (b) we have, respectively, the rectangles that contribute towards
the lower estimate, L(P), and the upper estimate, U (P), of A based on the partition P.
Note that, for clarity, only the first, second and last rectangles are shown.
55
3
3. The Riemann integral
That is, we can see that using these estimates we have
L(P) < A < U (P),
3
and so these estimates provide us with bounds on the true value of A. Indeed, since it is
easy to work out the area of a rectangle using ‘base times height’, we can see that the
base of each of the rectangles is given by the length of the relevant sub-interval, i.e. for
i = 1, 2, . . . , n we have a
sub-interval [xi−1 , xi ] with a length of xi − xi−1 ,
which means that, thinking about the heights as well, we have a
lower estimate given by
L(P) =
n
X
i=1
(xi − xi−1 )mi ,
where mi is the minimum value of f (x) for values of x in the sub-interval [xi−1 , xi ],
i.e.
mi = min{f (x)|xi−1 ≤ x ≤ xi },
is the height of this rectangle.
upper estimate given by
U (P) =
n
X
i=1
(xi − xi−1 )Mi ,
where Mi is the maximum value of f (x) for values of x in the sub-interval [xi−1 , xi ],
i.e.
Mi = max{f (x)|xi−1 ≤ x ≤ xi },
is the height of this rectangle.
Let’s have a look at a simple example to see how these estimates can be found in
practice.
Example 3.1 Suppose that we have the function f (x) = 1 + x2 defined over the
interval [−2, 2] and the partition P = {−2, 0, 1, 2} of this interval. Find the lower
and upper estimates, L(P) and U (P) respectively, of the area of the region bounded
by the curve y = f (x), the x-axis and the vertical lines x = −2 and x = 2.
If we sketch the curve y = 1 + x2 over the interval [−2, 2] and indicate the points in
the partition P = {−2, 0, 1, 2} we see that we will be looking at three rectangles
whose bases are given by the sub-intervals [−2, 0], [0, 1] and [1, 2]. Indeed, for the
lower estimate, we need to sum the areas of the three rectangles illustrated in
Figure 3.3(a) where the height of each rectangle is given by the minimum value
of f (x) in each of the sub-intervals. This gives us,
L(P) = (2)(1) + (1)(1) + (1)(2) = 5,
where we have taken each of the sub-intervals in turn and used ‘base times
height’ to find the area of each rectangle.
56
3.1. The Riemann integral
upper estimate, we need to sum the areas of the three rectangles illustrated in
Figure 3.3(b) where the height of each rectangle is given by the maximum value
of f (x) in each of the sub-intervals. This gives us,
U (P) = (2)(5) + (1)(2) + (1)(5) = 17,
where we have taken each of the sub-intervals in turn and used ‘base times
height’ to find the area of each rectangle.
3
In particular, observe that if A is the area of the region bounded by the curve
y = 1 + x2 , the x-axis and the vertical lines x = −2 and x = 2, then we can see from
the illustrations in Figure 3.3 that we have
L(P) < A < U (P),
as we should expect from the lower and upper estimates respectively.
y
y
y = 1 + x2
−2
2
2
1
1
O
(a)
1
2
x
y = 1 + x2
5
−2
O
1
x
2
(b)
Figure 3.3: In (a) and (b) we have, respectively, the rectangles that contribute towards
the lower estimate, L(P), and the upper estimate, U (P), of A based on the partition P.
Activity 3.1 Use integration to find the area, A, of the region bounded by the
curve y = 1 + x2 , the x-axis and the vertical lines x = −2 and x = 2. Hence verify
that the lower and upper estimates found in Example 3.1 do indeed satisfy the
inequality L(P) < A < U (P).
Of course, although we can find lower and upper estimates using simple partitions like
the one in Example 3.1, it will be necessary for us to use more sophisticated partitions
if we want to make any real progress towards more accurate estimates and, ultimately, a
definition of the Riemann integral.
57
3. The Riemann integral
Example 3.2 Suppose that we have the function f (x) = 1 + x defined over the
interval [0, 1] and the partitions
,1 ,
Pn = 0, n1 , n2 , . . . , n−1
n
3
of this interval for some n ∈ N. Find the lower and upper estimates, L(Pn ) and
U (Pn ) respectively, of the area of the region bounded by the curve y = f (x), the
x-axis and the vertical lines x = 0 and x = 1.
If we sketch the curve y = 1 + x over the interval [0, 1] and indicate the points in the
partition Pn , we see
we will be looking at n rectangles whose bases are given by
that
k
,
for k = 1, 2, . . . , n. Indeed, we see that, for k = 1, 2, . . . , n,
the sub-intervals k−1
n
n
each of these sub-intervals gives us a base whose length is
1
k k−1
−
= ,
n
n
n
and so, for the
lower estimate, we need to sum the areas of the n rectangles illustrated in
Figure 3.4(a) where the height of each rectangle is given by the minimum value
of f (x) in each of the sub-intervals. This gives us,
1
0
1
1
1
2
1
n−1
L(Pn ) =
1+
+
1+
+
1+
+ ··· +
1+
,
n
n
n
n
n
n
n
n
where we have taken each of the sub-intervals in turn and used ‘base times
height’ to find the area of each rectangle. Then, summing this series we find that
1
1
3−
,
L(Pn ) =
2
n
as you can verify in Activity 3.2.
upper estimate, we need to sum the areas of the n rectangles illustrated in
Figure 3.4(b) where the height of each rectangle is given by the maximum value
of f (x) in each of the sub-intervals. This gives us,
1
1
1
2
1
3
1
n
U (Pn ) =
1+
+
1+
+
1+
+ ··· +
1+
,
n
n
n
n
n
n
n
n
where we have taken each of the sub-intervals in turn and used ‘base times
height’ to find the area of each rectangle. Then, summing this series we find that
1
1
3+
,
U (Pn ) =
2
n
as you can verify in Activity 3.2.
In particular, observe that if A is the area of the region bounded by the curve
y = f (x), the x-axis and the vertical lines x = 0 and x = 1, then we can see from the
illustrations in Figure 3.4 that we have
L(Pn ) < A < U (Pn ),
as we should expect from the lower and upper estimates respectively.
58
3.1. The Riemann integral
y
y
y =1+x
2
1
y =1+x
2
3
1
...
O
1
n
2
n
...
3
n
...
n−1
n
1
x
O
(a)
1
n
2
n
...
3
n
n−1
n
1
x
(b)
Figure 3.4: In (a) and (b) we have, respectively, the rectangles that contribute towards
the lower estimates, L(Pn ), and the upper estimates, U (Pn ), of A based on the partition
Pn . Note that, for clarity, some of the y-intercepts have been omitted and only four of
the rectangles are shown.
Activity 3.2
formula
Given that the sum of the first k natural numbers is given by the
k
(k + 1),
2
show that the lower and upper estimates in Example 3.2 can be written as
1
1
1
1
L(Pn ) =
3−
and
U (Pn ) =
3+
,
2
n
2
n
1 + 2 + ··· + k =
when the series are summed.
Activity 3.3 Find the area, A, of the region bounded by the curve y = 1 + x, the
x-axis and the vertical lines x = 0 and x = 1. Hence verify that the lower and upper
estimates found in Example 3.2 do indeed satisfy the inequality L(Pn ) < A < U (Pn ).
Another thing to notice from Example 3.2 is that, as the function is increasing, there is
a straightforward relationship between the height of each rectangle and the value of the
function at the end-points of each sub-interval. In particular, we see that when we have
an increasing function over an interval, we find the
lower estimate takes the height of each rectangle to be the value of the function
at the left-hand end-point of the relevant sub-interval.
upper estimate takes the height of each rectangle to be the value of the function
at the right-hand end-point of the relevant sub-interval.
This idea is explored in Activity 3.4.
59
3. The Riemann integral
Activity 3.4 Suppose that f (x) is a non-negative continuous function defined over
some interval [a, b] and that P is the partition
{x0 , x1 , x2 , . . . , xn−1 , xn } where a = x0 < x1 < x2 < · · · < xn = b.
3
If f (x) is an increasing function on the interval [a, b], show that the lower and upper
estimates, L(P) and U (P) respectively, of the area of the region bounded by the
curve y = f (x), the x-axis and the vertical lines x = a and x = b are given by
L(P) =
n
X
i=1
(xi − xi−1 )f (xi−1 ) and U (P) =
n
X
i=1
(xi − xi−1 )f (xi ).
What are L(P) and U (P) if f (x) is a decreasing function on the interval [a, b]?
However, even though we can come up with such useful results about the lower and
upper estimates of an area for increasing or decreasing functions over an interval,
generally speaking, the best way to proceed is to sketch the relevant rectangles and then
use these to infer their areas as we did in Examples 3.1 and 3.2.
3.1.2
Getting better lower and upper estimates
Having seen how to find lower and upper estimates of an area, we now want to see how
we can get better lower and upper estimates with a view to finding the best lower and
upper estimates which should, naturally, be the value of the area itself. In particular, as
the next example makes clear, we should be able to get better lower and upper
estimates by adding points to our partition so that we get ‘thinner’ rectangles and hence
better estimates of the area involved.
Example 3.3 Use the partition P 0 = {−2, −1, 0, 1, 2} to find the lower and upper
estimates, L(P 0 ) and U (P 0 ) respectively, of the area discussed in Example 3.1.
Compare these values to what we found in that example.
If we sketch the curve y = 1 + x2 over the interval [−2, 2] and indicate the points in
the partition P 0 = {−2, −1, 0, 1, 2} we see that we will now be looking at four
rectangles whose bases are given by the sub-intervals [−2, −1], [−1, 0], [0, 1] and
[1, 2]. Indeed, for the
lower estimate, we need to sum the areas of the four rectangles illustrated in
Figure 3.5(a) where the height of each rectangle is given by the minimum value
of f (x) in each of the sub-intervals. This gives us,
L(P 0 ) = (1)(2) + (1)(1) + (1)(1) + (1)(2) = 6,
where we have taken each of the sub-intervals in turn and used ‘base times
height’ to find the area of each rectangle.
upper estimate, we need to sum the areas of the four rectangles illustrated in
Figure 3.5(b) where the height of each rectangle is given by the maximum value
60
3.1. The Riemann integral
of f (x) in each of the sub-intervals. This gives us,
U (P 0 ) = (1)(5) + (1)(2) + (1)(2) + (1)(5) = 14,
where we have taken each of the sub-intervals in turn and used ‘base times
height’ to find the area of each rectangle.
3
Comparing this with our answers from Example 3.1, we can see that the
lower estimates give us
L(P) = 5 < 6 = L(P 0 )
=⇒
L(P) < L(P 0 ),
i.e. L(P 0 ) is a larger, and hence better, lower estimate of the area than L(P).
upper estimates give us
U (P 0 ) = 14 < 17 = U (P)
=⇒
U (P 0 ) < U (P),
i.e. U (P 0 ) is a smaller, and hence better, upper estimate of the area than U (P).
Indeed, it should be clear that this improvement in the lower and upper estimates
has come about because P 0 has one more point than P and this has enabled us to
replace the single rectangle for the sub-interval [−2, 0] from P (see Figure 3.3) with
two ‘thinner’ rectangles for the sub-intervals [−2, −1] and [−1, 0] from P 0 (see
Figure 3.5).
In particular, observe that if A is the area of the region bounded by the curve
y = 1 + x2 , the x-axis and the vertical lines x = −2 and x = 2, then we can see from
the results above and the illustrations in Figure 3.5 that we have
L(P) < L(P 0 ) < A < U (P 0 ) < U (P),
i.e. P 0 gives us better lower and upper estimates of A than P.
Activity 3.5 Use your answer to Activity 3.1 to verify that the lower and upper
estimates found in Example 3.2 satisfy the inequality
L(P) < L(P 0 ) < A < U (P 0 ) < U (P),
as one would expect.
In fact, this is what we should expect more generally, i.e.
lower (and upper) estimates should always be less (or greater) than the area they
are estimates of, and
increasing the number of the points in the partition should always give us better
lower (or upper) estimates.
To make this clear, let’s consider another example.
61
3. The Riemann integral
y
y
y = 1 + x2
y = 1 + x2
5
3
−2
2
2
1
1
O
−1
1
2
x
−2
(a)
−1
O
1
2
x
(b)
Figure 3.5: In (a) and (b) we have, respectively, the rectangles that contribute towards
the lower estimate, L(P 0 ), and the upper estimate, U (P 0 ), of A based on the partition P.
Example 3.4
Use the results from Example 3.2 to verify the two points above.
We start by noting that in this case, as we saw in Activity 3.3, we can meaningfully
talk about the value of the area, A, of the region bounded by the curve y = 1 + x,
the x-axis and the vertical lines x = 0 and x = 1 because it is simply the area of a
trapezium. Indeed, for the first point, we have already shown in Activity 3.3 that
L(Pn ) < A < U (Pn ),
i.e. if we are estimating the area A, then whatever value of n ∈ N we take, the lower
estimates, L(Pn ), are always less than A whereas the upper estimates, U (Pn ) are
always greater than A.
For the second point, consider that the partition, Pn has n + 1 points and, if we take
some other natural number m > n, the partition Pm will have m + 1 points, i.e. Pm
has m − n more points than Pn . We then have
lower estimates given by
1
1
L(Pn ) =
3−
2
n
and
1
L(Pm ) =
2
1
3−
.
m
But, m > n > 0 means that
1
1
>
n
m
=⇒
1
1
3− <3−
n
m
=⇒
1
2
1
3−
n
1
<
2
1
3−
m
,
which means that we have L(Pn ) < L(Pm ). Indeed, using the first point, both of
these lower estimates must be less than A, and so we have
L(Pn ) < L(Pm ) < A,
i.e. increasing the number of points in the partition has given us a better lower
estimate.
62
3.1. The Riemann integral
upper estimates given by
1
1
3+
U (Pn ) =
2
n
and
1
U (Pm ) =
2
1
3+
m
.
But, m > n > 0 means that
1
1
>
n
m
=⇒
1
1
3+ >3+
n
m
=⇒
1
2
1
3+
n
1
>
2
1
3+
m
3
,
which means that we have U (Pn ) > U (Pm ). Indeed, using the first point, both
of these upper estimates must be greater than A, and so we have
U (Pn ) > U (Pm ) > A,
i.e. increasing the number of points in the partition has given us a better upper
estimate.
Notice, however, that although we have been using various graphical and geometric
‘intuitions’ about the area under a curve in our examples, we have not actually defined
what the area under a curve is. In particular, this is why all reference to using definite
integrals to find areas have been relegated to Activities 3.1 and 3.3 where they can serve
to illustrate the points being made without interfering with the flow of our argument.
3.1.3
The definition of the Riemann integral
We now turn to the definition of the Riemann integral which will allow us to assign a
value to the definite integral
Z b
f (x) dx,
a
where f (x) is a non-negative continuous function over the interval [a, b] by taking it to
be the value of the area of the region between the curve y = f (x), the x-axis and the
vertical lines x = a and x = b. So, we have to ask ourselves, what is the value of this
area?
It should be clear, from what we have seen above, that in the case we are considering:
Every partition of [a, b] gives us a lower and upper estimate of the area.
Every lower estimate is less than every upper estimate, i.e. we have L(P) ≤ U (P).
This means that there is at least one number A such that L(P) ≤ A ≤ U (P).
Now, if there is only one such number, then A is the value of the area and hence the
value of the Riemann integral
Z b
f (x) dx,
a
whereas if there is more than one such number, then we say that the area is undefined
and so this Riemann integral is undefined too.
63
3. The Riemann integral
In practice, we can use this to find the value of the Riemann integral,
Z b
f (x) dx,
I=
a
3
where f (x) is a non-negative continuous function as follows. We find the lower and
upper estimates in terms of some appropriately general partition, P, and then find the
smallest number, L∗ , that is greater than all the lower estimates, L(P), and the
largest number, U ∗ , that is smaller than all the upper estimates, U (P).
Then, according to what we have seen above, we should find that
L∗ ≤ I ≤ U ∗ ,
so that, if there is only one number, I, that satisfies this inequality we can take this to
be the value of the Riemann integral.2
Example 3.5
Continuing on from Examples 3.2 and 3.4, find the value of
Z 1
(1 + x) dx,
0
using the definition of the Riemann integral.
We saw in Example 3.2 that we have
lower estimates given by
1
L(Pn ) =
2
1
3−
n
3
≤ .
2
for all n ∈ N. That is, following on from Example 3.4, we have found that, for
m, n ∈ N with m > n,
L(Pn ) ≤ L(Pm ) ≤ L∗ ,
where L∗ = 3/2 is the smallest number that is greater than all the lower
bounds.3
upper estimates given by
1
U (Pn ) =
2
1
3
3+
≥ ,
n
2
2
Technically, we are looking at every possible partition, P, of the interval [a, b] and, having found the
corresponding lower and upper estimates, we take L∗ to be the least upper bound (or supremum) of the
lower estimates and U ∗ to be the greatest lower bound (or infimum) of the upper estimates, i.e.
L∗ = sup{L(P)|P is a partition of [a, b]} and U ∗ = inf{U (P)|P is a partition of [a, b]}.
This is because, generally, as we have taken a partition to be a finite set of points, we shall find that
there is no partition P that makes L∗ = L(P) or U ∗ = U (P), even though L∗ and U ∗ act as the relevant
bounds on L(P) and U (P) respectively.
64
3.1. The Riemann integral
for all n ∈ N. That is, following on from Example 3.4, we have found that, for
m, n ∈ N with m > n,
U ∗ ≤ U (Pm ) ≤ U (Pn ),
where U ∗ = 3/2 is the largest number that is smaller than all the upper bounds.
This means that there is only one number, I, that satisfies the inequality
3
L∗ ≤ I ≤ U ∗ ,
in this case, i.e. I = 3/2, and so we take this to be the value of the given Riemann
integral. That is,
Z 1
3
(1 + x) dx = ,
2
0
as we should expect from Activity 3.3.
Activity 3.6
Use the definition of the Riemann integral to find the value of
Z 2
(1 + x2 ) dx,
0
by considering, for m ∈ N, the partition
, 2},
Pm = {0, m1 , m2 , . . . , 2m−1
m
of the interval [0, 2].
Note: You will need to use the formula
12 + 22 + · · · + k 2 =
k
(k + 1)(2k + 1),
6
for the sum of the first k square numbers.
Of course, more generally, we should expect the Riemann integral
Z b
f (x) dx,
a
to be well-defined if f (x) is a non-negative continuous function over the interval [a, b].
This is because it is just the area between the curve y = f (x), the x-axis and the
vertical lines x = a and x = b, which is clearly well-defined if f (x) is continuous. Indeed,
although we do not dwell on this here, we can drop the requirement that f (x) is
non-negative and extend what we have seen to functions where f (x) is negative for
some values of x in the interval [a, b] by utilising the methods discussed in Section 5.3 of
174 Calculus. All of this leads to the following fact.
3
That is, we have
L∗ = sup{L(Pn )|n ∈ N},
as, for all n ∈ N, L(Pn ) ≤ L∗ and so L∗ is an upper bound on the lower estimates and L∗ is the least
upper bound as there is no other upper bound, l, such that l < L∗ . Also observe that, for any n ∈ N,
L(Pn ) < L∗ and so there is no partition with a finite number of points that makes L(Pn ) = L∗ .
65
3. The Riemann integral
If f (x) is a continuous function over the interval [a, b], then the Riemann
integral
Z b
f (x) dx,
a
exists.
3
Although, we shall not prove this here.
3.1.4
What happens if the integrand isn’t continuous?
We now begin to consider the Riemann integral
Z b
f (x) dx,
a
when f (x) is not continuous over the interval [a, b]. In particular, we consider how the
definition needs to be slightly modified when the integrand has finite discontinuities at
certain points and encounter a case where the Riemann integral is undefined.
In particular, one problem we will encounter is that a function which has finite
discontinuities may not have a minimum or maximum value over every interval. To
motivate the discussion that follows, let’s look at an example.
Example 3.6
Consider the integral
Z
(
1
f (x) dx where f (x) =
0
2x if 0 ≤ x < 1
1
if x = 1
and, for n ∈ N, the partition
Pn = {0, n1 , n2 , . . . , 1},
of the interval [0, 1]. Sketch f (x) over the interval [0, 1] and explain why the upper
estimates (as we defined them in Section 3.1.1) are problematic in this case. Does
this seem reasonable?
If we sketch f (x) over the interval [0, 1], as illustrated in Figure 3.6(a), we see that
the function has a finite discontinuity at x = 1 because f (x) → 2 as x → 1− but
f (1) = 1 6= 2. In particular, even though f (x) → 2 as x → 1− , the function never
attains the value two and so, over the interval [0, 1], this function has no maximum
value.4
This poses a problem for the upper estimates since, if we consider the sub-interval
[ n−1
, 1], as illustrated in Figure 3.6(b), it means that we can not ascribe a value to
n
Mn = max{f (x)| n−1
≤ x ≤ 1},
n
and so, bearing in mind that the base of each rectangle that arises from the partition
is 1/n, the upper estimates
n
X
1
Mi ,
U (Pn ) =
n
i=1
66
3.1. The Riemann integral
for n ∈ N are not defined since we cannot ascribe a value to its last term.
However, given that we want to interpret the given Riemann integral as the area of
the region under the curve y = f (x), the x-axis and the vertical lines x = 0 and
x = 1, we would surely expect this area to exist. Indeed, we would expect this to be
the area of a triangular region which, using ‘half base times height’, is one! So, in
this case, it seems unreasonable that the Riemann integral does not exist because we
can not define its upper estimates.
y
y
y = f (x)
2
y = f (x)
2
1
1
O
O
x
1
(a)
n−1
n
1
x
(b)
Figure 3.6: In (a) we have a sketch of the function y = f (x) from Example 3.6 and in (b)
, 1].
we examine the effect of the problematic sub-interval [ n−1
n
To deal with cases like this, we should use a slightly more general definition of the
Riemann integral
Z b
f (x) dx,
a
in terms of a partition, P, of the interval [a, b] given by
P = {x0 , x1 , x2 , . . . , xn−1 , xn } where a = x0 < x1 < x2 < · · · < xn = b,
which runs as before except that, now, we have
lower estimates, L(P), given by
L(P) =
n
X
i=1
(xi − xi−1 )mi ,
where mi is the greatest lower bound, or infimum, of f (x) for values of x in the
sub-interval [xi−1 , xi ], i.e.
mi = inf{f (x)|xi−1 ≤ x ≤ xi },
4
Of course, the function g(x) = 2x defined over the interval [0, 1] gives us g(1) = 2 and this is its
maximum value, but the function f (x) that we are considering here has f (1) = 1!
67
3
3. The Riemann integral
with the understanding that, in cases where this value is actually attained by the
function in the sub-interval, this will just be the minimum value of the function as
before.
upper estimates, U (P), given by
U (P) =
3
n
X
i=1
(xi − xi−1 )Mi ,
where Mi is the supremum of f (x) for values of x in the sub-interval [xi−1 , xi ], i.e.
Mi = sup{f (x)|xi−1 ≤ x ≤ xi },
with the understanding that, in cases where this value is actually attained by the
function in the sub-interval, this will just be the maximum value of the function as
before.
Of course, with this in mind, we can now make sense of what we saw in Example 3.6 as
the next example shows.
Example 3.7 Following on from Example 3.6, use this more general definition of
the Riemann integral to find the value of
(
Z 1
2x if 0 ≤ x < 1
f (x) dx where f (x) =
1
if x = 1
0
by considering, for n ∈ N with n ≥ 2, the partitions
Pn = {0, n1 , n2 , . . . , 1},
of the interval [0, 1].
If we sketch the curve y = f (x) over the interval [0, 1] and indicate the points in the
partition Pn we see
that
we will be looking at n rectangles whose bases are given by
k
the sub-intervals k−1
,
for k = 1, 2, . . . , n. Indeed, we see that, for k = 1, 2, . . . , n,
n
n
each of these sub-intervals gives us a base whose length is
1
k k−1
−
= ,
n
n
n
and so, for the
lower estimates, we need to sum the areas of the n rectangles illustrated in
Figure 3.7(a) where the height of each rectangle is given by the minimum value
of f (x) in each of the sub-intervals. This gives us,
1
0
1
1
1
2
1
n−2
1
L(Pn ) =
2·
+
2·
+
2·
+ ··· +
2·
+
1 ,
n
n
n
n
n
n
n
n
n
where we have taken each of the sub-intervals in turn and used ‘base times
height’ to find the area of each rectangle. Then, summing this series we find that
n−1
≤ 1,
n2
for n ≥ 2, as you can verify in Activity 3.7. That is, L∗ = 1 is the smallest
number that is greater than all the lower bounds.
L(Pn ) = 1 − 2 ·
68
3.1. The Riemann integral
upper estimates, we need to sum the areas of the n rectangles illustrated in
Figure 3.7(b) where the height of each rectangle is given by the maximum value
of f (x) in the first n − 1 sub-intervals and the supremum of f (x) in the last
sub-interval. This gives us,
1
1
1
2
1
3
1
n−1
1
n
U (Pn ) =
2·
+
2·
+
2·
+···+
2·
+
2·
,
n
n
n
n
n
n
n
n
n
n
where we have taken each of the sub-intervals in turn and used ‘base times
height’ to find the area of each rectangle. Then, summing this series we find that
U (Pn ) = 1 +
1
≥ 1,
n
for n ≥ 2, as you can verify in Activity 3.7. That is, U ∗ = 1 is the largest
number that is less than all the upper bounds.
This means that there is only one number, I, that satisfies the inequality
L∗ ≤ I ≤ U ∗ ,
in this case, i.e. I = 1, and so we take this to be the value of the given Riemann
integral. That is,
(
Z 1
2x if 0 ≤ x < 1,
f (x) dx = 1 when f (x) =
1
if x = 1,
0
as we should expect from our discussion in Example 3.6.
Activity 3.7 Using the formula for the sum of the first k natural numbers given in
Activity 3.2, show that the lower and upper estimates in Example 3.7 can be written
as
1
n−1
L(Pn ) = 1 − 2 ·
and U (Pn ) = 1 + ,
2
n
n
when the series are summed and verify that, for n ≥ 2, we have L(Pn ) ≤ 1 and
U (Pn ) ≥ 1.
However, if there are too many finite discontinuities, the Riemann integral will not exist
as the next example shows.
Example 3.8
Consider the integral
Z
(
1
f (x) dx where f (x) =
0
1 if x is rational,
0 if x is irrational,
and the partition
P = {x0 , x1 , x2 , . . . , xn−1 , xn } where 0 = x0 < x1 < x2 < · · · < xn−1 < xn = 1,
69
3
3. The Riemann integral
y
y
y = f (x)
2
3
1
y = f (x)
2
1
...
O
1
n
2
n
3
n
...
...
n−2
n
n−1
n
1
x
O
1
n
(a)
2
n
3
n
...
n−2
n
n−1
n
1
x
(b)
Figure 3.7: In (a) and (b) we have, respectively, the rectangles that contribute towards
the lower estimates, L(Pn ), and the upper estimates, U (Pn ), of the Riemann integral in
Example 3.7 based on the partition Pn for n ≥ 2. Note that, for clarity, some of the
y-intercepts have been omitted and only five of the rectangles are shown. (Of course, in
(a), the first rectangle has a height of zero and so we can not see it!)
of the interval [0, 1]. Find the lower and upper estimates, L(P) and U (P)
respectively, of this integral. What do these tell you about the existence of this
Riemann integral?
Note: A real number is said to be rational if it is of the form p/q where p, q ∈ Z and
q 6= 0 and, if it is not rational, a number is said to be irrational.5 Indeed, we will
need the facts that
(1) every interval [a, b] with a < b contains at least one rational number, and
(2) every interval [a, b] with a < b contains at least one irrational number,
to answer this question.
In this case, we can not sketch the function as it is discontinuous at every point, but
we can easily find the
lower estimate by noting that in every sub-interval [xi−1 , xi ] of length
xi − xi−1 the minimum value of f (x) is zero as, by fact (1), there is at least one
rational number in that sub-interval. This gives us
L(P) = (x1 − x0 )(0) + (x2 − x1 )(0) + · · · + (xn−1 − xn−2 )(0) + (xn − xn−1 )(0) = 0,
i.e. the lower estimate is zero for any partition, P.
upper estimate by noting that in every sub-interval [xi−1 , xi ] of length
xi − xi−1 the maximum value of f (x) is one as, by fact (2), there is at least one
irrational number in that sub-interval. This gives us
U (P) = (x1 −x0 )(1)+(x2 −x1 )(1)+· · ·+(xn−1 −xn−2 )(1)+(xn −xn−1 )(1) = xn −x0 ,
70
3.1. The Riemann integral
if we cancel the corresponding intermediate terms and, noting that x0 = 0 and
xn = 1, we then see that the upper estimate is one for any partition, P.
Consequently, we see that as L(P) = 0 and U (P) = 1 for any partition P, there are
many numbers I that satisfy the inequality
L(P) ≤ I ≤ U (P),
and so, as there is not only one such number, we can not assign a single value to this
Riemann integral, i.e. it does not exist.6
In the next chapter, we will see how to deal with Riemann integrals in cases where the
interval we are concerned with is not finite and where the integrand fails to be
continuous in other, less straightforward, ways (such as the presence of an infinite
discontinuity).
3.1.5
Some properties of the Riemann integral
So far, we have been concerned with Riemann integrals of the form
Z b
f (x) dx,
a
where a < b and we now want to extend our understanding of such integrals by seeing
what happens when a = b or a > b. To do this, we introduce some basic properties of
the Riemann integral in Theorem 3.1 and, in particular, these will be useful in the next
section.
Theorem 3.1 If the Riemann integral of f exists over some interval containing the
points a, b and c with a ≤ c ≤ b, then
(a) The integral over an interval of zero length is zero, i.e.
Z a
f (x) dx = 0.
a
(b) The integral is additive over the interval of integration, i.e.
Z b
Z c
Z b
f (x) dx =
f (x) dx +
f (x) dx.
a
a
c
(c) Interchanging the limits of integration changes the sign of the integral, i.e.
Z b
Z a
f (x) dx = −
f (x) dx.
a
b
√
For example, you may have encountered the fact that the real number, 2, is irrational because it
can not be written in the form p/q where p, q ∈ Z and q 6= 0.
6
Actually, there is a well-defined sense in which ‘almost all’ real numbers are irrational and so the
value of this integral ‘ought’ to be zero as its value shouldn’t be affected by the value of the integrand
at a ‘few’ points where its argument is rational. Indeed, although it is beyond the scope of this course,
you might care to note that there is a more general notion of integration, called the Lebesgue integral,
whose value would be zero in this case.
5
71
3
3. The Riemann integral
(d) It follows that
Z
b
Z
b
f (x) dx −
f (x) dx =
a
c
Z
a
f (x) dx.
c
We will not prove this theorem here but, in Activity 3.8, you will be able to motivate
these results by appealing to our interpretation of the Riemann integral as an area.
3
Activity 3.8 Suppose that f is a non-negative function whose Riemann integral
exists over some interval containing the points a, b and c. By considering the areas
involved, explain why parts (a) and (b) of Theorem 3.1 hold. Hence deduce parts (c)
and (d) of this theorem.
3.2
The Fundamental Theorem of Calculus
Returning to the case where the function, f (x), is continuous and non-negative over an
interval [a, b], we have defined the Riemann integral,
Z
b
f (x) dx,
a
in terms of the area between the curve y = f (x), the x-axis and the vertical lines x = a
and x = b. What we want to do now is make sure that this view of integration agrees
with what we saw in Chapter 5 of 174 Calculus. In particular, we want to be sure that
the relationship between integrals and derivatives that we used there still works if
integration is defined in this way. Indeed, the key result here is the Fundamental
Theorem of Calculus (which we will abbreviate to FTC) and this runs as follows.
Theorem 3.2 (FTC) If f is a continuous function defined on the interval [a, b], then
Z t
d
f (x) dx = f (t),
dt
c
for any numbers c and t in the interval (a, b).
We will spend the rest of this chapter seeing why this works and what it tells us about
the relationship between integrals and derivatives.
3.2.1
Motivating the FTC
To see how the FTC arises, consider the case where f (x) is a decreasing non-negative
continuous function over the interval [a, b]. In particular, we will take c, t ∈ (a, b) where
x > c and consider the integral
Z t
A(t) =
f (x) dx,
c
so that A(t) represents the area between the curve y = f (x), the x-axis and the vertical
lines x = c and x = t as illustrated in Figure 3.8(a). Indeed, this means that if h > 0, we
72
3.2. The Fundamental Theorem of Calculus
also have the integral
t+h
Z
A(t + h) =
f (x) dx,
c
so that A(t) represents the area between the curve y = f (x), the x-axis and the vertical
lines x = c and x = t + h as illustrated in Figure 3.8(b).
y
3
y
y = f (x)
y = f (x)
A(t)
O
a
A(t + h)
c
t
x
b
O
a
x
t t+h b
c
(a)
(b)
Figure 3.8: The shaded regions in (a) and (b) have areas A(t) and A(t + h) respectively.
We now consider the difference between these two areas, i.e. the quantity
A(t + h) − A(t),
which is the area of the vertically hatched region indicated in Figure 3.9.
y
O
y
a
c
1111
010000
101111
0000
101111
0000
101111
0000
101111
0000
10
t t+h b
(a)
y = f (x)
x
O
a
c
1111
0000
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
t t+h b
y = f (x)
x
(b)
Figure 3.9: The areas of the shaded regions in (a) and (b) are the lower and upper
estimates, respectively, of the area A(t + h) − A(t) of the vertically hatched region.
73
3. The Riemann integral
In particular, we see that since f (x) is a decreasing function over the interval [a, b], we
have a
lower estimate of this area given by the area of the shaded rectangle in
Figure 3.9(a) which, using ‘base times height’, is
hf (t + h),
3
and so we have hf (t + h) ≤ A(t + h) − A(t).
upper estimate of this area given by the area of the shaded rectangle in
Figure 3.9(b) which, using ‘base times height’, is
hf (t),
and so we have A(t + h) − A(t) ≤ hf (t).
Consequently, putting these two inequalities together, we have
hf (t + h) ≤ A(t + h) − A(t) ≤ hf (t)
=⇒
f (t + h) ≤
A(t + h) − A(t)
≤ f (t),
h
as h > 0. Then, we take the limit as h → 0+ on both sides of this inequality to get
lim+ f (t+h) ≤ lim+
h→0
h→0
A(t + h) − A(t)
≤ lim+ f (t)
h→0
h
=⇒
f (t) ≤ lim+
h→0
A(t + h) − A(t)
≤ f (t),
h
if we use the continuity of f (x) over [a, b].7 This means that
lim+
h→0
A(t + h) − A(t)
= f (t),
h
and, repeating the argument with h < 0, we also get
lim−
h→0
A(t + h) − A(t)
= f (t),
h
as you can verify in Activity 3.9. Then, putting these two results together, we have
A(t + h) − A(t)
= f (t)
h→0
h
lim
=⇒
A0 (t) = f (t),
if we use the definition of the derivative, A0 (t), of A(t). Thus, we have shown that when
f (x) is a continuous decreasing function over [a, b] with c, t ∈ (a, b) and c < t, we have
Z t
d
0
A (t) =
f (x) dx = f (t),
dt
c
as expected from the FTC.
7
Notice that the continuity of f (x) over [a, b] is essential here in order to justify the assertion that
the limit
lim f (t + h),
h→0+
exists and that it is equal to f (t). Also, notice that we can only take the limit as h → 0+ here as we
have assumed that h > 0.
74
3.2. The Fundamental Theorem of Calculus
Activity 3.9
Show that
lim−
h→0
A(t + h) − A(t)
= f (t),
h
using an argument similar to the one above with h < 0.
3
3.2.2
Notation: Dummy variables
Before we proceed, we should make sure that we understand the role that dummy
variables are playing here. Generally, given a function, f , defined over some interval
[a, b] it makes no difference what we call the independent variable. That is, we could use
x and look at values of
f (x) where x ∈ [a, b],
or we could use y and look at values of
f (y) where y ∈ [a, b],
or we could use any other letter to represent the independent variable.8 Indeed, when
we look at an integral like
Z b
f (x) dx,
a
where a and b are constants, we know that the answer will not depend on x and, in
particular, it makes no difference if we write this integral as
Z b
f (y) dy.
a
In such cases, the variable that is used inside the integral is a dummy variable as it does
not matter what we call it since it does not occur in the final answer. However, when we
have an integral like
Z
t
f (x) dx,
c
where only c is a constant, we now have two variables. One of these variables, x, is a
dummy variable as it will not figure in the final answer but the other variable, t, is not
as the answer will be a function of t, i.e. we may write
Z t
F (t) =
f (x) dx.
c
Of course, this means that
Z
t
Z
f (x) dx and
c
t
f (y) dy,
c
say, represent the same integral whereas
Z t
Z s
F (t) =
f (x) dx and F (s) =
f (x) dx,
c
c
8
Although, by convention, we usually use letters from the end of the English alphabet to denote
variables.
75
3. The Riemann integral
say, represent the same function, namely F . In particular, this means that some care
must be taken when choosing names for our variables as we would not want to write
Z x
f (x) dx,
F (x) =
c
3
because this uses x as both the independent variable of F and the dummy variable of
the integral. To see why this is a problem, ask yourself what is, say, F (1) using this
definition of F ? Is it
Z
Z
1
1
f (x) dx or
f (1) d1 ?
c
c
Obviously, we want to be able to distinguish between these two options and that is why
we use dummy variables!
3.2.3
The relationship between integration and differentiation
Having motivated the FTC, we are now in a position to clarify the relationship between
integration and differentiation. That is, we want to find that differentiation ‘undoes’
integration and that, up to an arbitrary constant, integration ‘undoes’ differentiation.
Differentiation ‘undoes’ integration
Recall that, in Section 5.1 of 174 Calculus, we defined an antiderivative of the function,
f (t), to be a function F (t) such that
dF
= f (t).
dt
Indeed, using the FTC, we can now see that the integral
Z t
f (x) dx,
F (t) =
c
where c is a constant is an antiderivative of f (t) since
Z t
dF
d
0
F (t) =
=
f (x) dx = f (t).
dt
dt
c
That is, differentiation ‘undoes’ (or ‘reverses’) the process of integration as, given the
integral of f , differentiation allows us to find f .
Integration ‘undoes’ differentiation
Conversely, if F (t) is an antiderivative of f (t), say we have
Z t
F (t) =
f (x) dx,
c
where c is a constant, as before, then for some constant d, we also have
Z t
Z t
Z d
f (x) dx =
f (x) dx −
f (x) dx = F (t) − F (d),
d
76
c
c
3.2. The Fundamental Theorem of Calculus
using Theorem 3.1(d) and, of course, F (d) is a constant. That is, integration ‘undoes’
(or ‘reverses’) the process of differentiation as, given the derivative of F , i.e. f ,
integration allows us to find F up to a constant. Indeed, we find that if F1 (t) and F2 (t)
are any two antiderivatives of f , we have
F10 (t) = f (t)
F20 (t) = f (t),
and
3
which means that
d
F1 (t) − F2 (t) = F10 (t) − F20 (t) = f (t) − f (t) = 0,
dt
and so the function F1 (t) − F2 (t) must be a constant as its derivative is zero.
Consequently, any two antiderivatives of f can only differ by a constant and so, as we
should expect, integration only ‘undoes’ differentiation up to an arbitrary constant.
In summary
Having established that the expected relationships between integration and
differentiation hold, we can now evaluate Riemann integrals in the manner described in
Chapter 5 of 174 Calculus, i.e. we can use our standard integrals and rules of
integration.
3.2.4
Some applications of the FTC
We are now in a position to consider some other applications of the FTC.
Using the FTC and Taylor series to approximate the value of an integral
At the beginning of this chapter, we observed that the integral
Z 1
2
e−x dx,
0
2
could not be found in the standard way as the integrand, e−x , has no antiderivative. Of
course, we could use our interpretation of the integral as an area to gain some
approximation to the value of this integral, but the FTC allows us to approximate its
value in another way as the next example shows.
Example 3.9
Given that
Z
F (t) =
t
2
e−x dx,
0
find the Taylor series for F (t) about t = 0 in terms up to t3 . Hence find an
approximate value for F (1).
To find the Taylor series for F (t) about t = 0, we note that
Z 0
Z t
2
−x2
F (t) =
e
dx =⇒ F (0) =
e−x dx = 0,
0
0
77
3. The Riemann integral
using Theorem 3.1(a) and that
Z t
d
2
−x2
0
e
dx = e−t
F (t) =
dt
0
=⇒
F 0 (0) = 1,
using the FTC. We then see that,
3
2
F 00 (t) = −2t e−t
=⇒
2
2
2
F 000 (t) = (−2)(e−t ) + (−2t)(−2t e−t ) = 2(2t2 − 1) e−t ,
which means that F 00 (0) = 0 and F 000 (0) = −2. Thus, we have
F (t) = F (0) + tF 0 (0) +
t2 00
t3
t3
F (0) + F 000 (0) + · · · = t − + · · · ,
2!
3!
3
as the Taylor series for F (t) about t = 0 in terms up to t3 . Consequently, we can see
that
2
1
F (1) ' 1 − = ,
3
3
is the required approximate value for F (1). Indeed, as
Z 1
2
F (1) =
e−x dx,
0
this gives us an approximate value of 0.67 (2dp) for this integral and this is fairly
close to the true value which turns out to be 0.75 (2dp).
Using the FTC to define functions in terms of definite integrals
We can also use the FTC to define certain functions in terms of definite integrals as the
next activity shows.
2
Activity 3.10 The function, F (x), satisfies the conditions F 0 (x) = e−x and
F (0) = 0. Write F (x) as a definite integral of the form
Z x
F (x) =
f (t) dt,
A
for a suitable choice of f and A.
Using the FTC to solve integral equations
We can also use the FTC to solve certain integral equations, i.e. equations that involve
definite integrals, as the next example shows.
Example 3.10
Use the FTC to solve the integral equation f (t) = 1 −
Z
t
f (x) dx.
0
If we differentiate both sides of this integral equation with respect to t, we get
f 0 (t) = −f (t),
78
3.2. The Fundamental Theorem of Calculus
where we have used the FTC to differentiate the integral. Solving this simple
separable differential equation in the standard way9 we get
f (t) = A e−t ,
for some constant A ∈ R. Indeed, setting t = 0 in the integral equation, we see that
Z 0
f (x) dx =⇒ f (0) = 1 − 0 = 1,
f (0) = 1 −
0
using Theorem 3.1(a), and so A = 1. Consequently, a function f that satisfies this
integral equation is
f (t) = e−t ,
and so this is the required solution.
In fact, we will see another way of solving integral equations like this in Chapter 7 when
we look at Laplace transforms.
3.2.5
An extension of the FTC
We can also use the FTC to differentiate more complicated integrals if we use the chain
rule. For instance, if we have the integral
Z q(t)
F (t) =
f (x) dx,
p(t)
we can use Theorem 3.1(d) to write it as
Z q(t)
Z
F (t) =
f (x) dx −
c
p(t)
f (x) dx,
c
where c is some constant. Then, differentiating both sides of this expression with
respect to t, we get
!
!
Z q(t)
Z p(t)
d
d
F 0 (t) =
f (x) dx −
f (x) dx ,
dt
dt
c
c
so that, using the chain rule, we have
Z q
Z p
dq d
dp d
0
F (t) =
f (x) dx −
f (x) dx .
dt dq
dt dp
c
c
Then, applying the FTC, this gives us
Z
d q(t)
0
F (t) =
f (x) dx = q 0 (t)f (q(t)) − p0 (t)f (p(t)),
dt p(t)
where, of course, this result assumes that p and q are both differentiable functions of t
and that f is continuous over the two intervals of integration. Let’s look at an example
to see how this works.
9
See, for example, Section 8.2.1 of 174 Calculus.
79
3
3. The Riemann integral
Example 3.11
Given that
Z
t3
F (t) =
x2 dx,
4t
0
find F (t) using the result above.
3
Here we have p(t) = 4t and q(t) = t3 , which means that
F 0 (t) = (3t2 )(t3 )2 − (4)(4t)2 = 3t8 − 64t2 ,
if we use the result above.
Activity 3.11 Following on from Example 3.11, use integration to find F (t) and
hence verify your answer for F 0 (t).
Indeed, we will see in Section 6.1.3 that this result can be extended even further when
we meet the Leibniz rule for differentiating integrals.
Using the FTC to derive identities
This extension of the FTC also has an interesting application because it can be used to
derive certain identities like the ones we saw in Section 2.1.4 of 174 Calculus.
Example 3.12
Use integration to verify that
Z t
1
dx,
ln t =
1 x
where t > 0. Hence show that if a > 0 is a constant and F is given by
F (t) = ln(at) − ln t,
where t > 0, then F 0 (t) is zero. Deduce that F (t) is a constant and, by finding this
constant, deduce that
ln(ab) = ln a + ln b,
where b > 0 is a constant too.
Note: Of course, if we use the fact that
ln(ab) = ln a + ln b
=⇒
ln(ab) − ln b = ln a,
for a, b > 0, then it is obvious that
F (t) = ln(at) − ln t = ln a,
for a, t > 0. Indeed, as a is a constant, this means that F (t) = ln a is a constant too
and so, obviously, F 0 (t) = 0! However, here we are using an integral that represents
ln x and the function F (t) to derive this fact and so we can not use it in this way
here.
80
3.2. The Fundamental Theorem of Calculus
Firstly, it is easy to verify that, for t > 0, we have
t
Z t
1
dx = ln x = ln t − ln 1 = ln t,
1 x
1
as ln 1 = 0. Then, for a, t > 0, we can use this to write
Z t
Z at
1
1
dx −
dx,
F (t) = ln(at) − ln t =
x
1 x
1
3
so that, using the FTC and its extension as appropriate, we have
1
1 1
1
0
− = − = 0,
F (t) = (a)
at
t
t
t
as required. But, of course, if F 0 (t) = 0, this means that F (t) = c where c is a
constant which we can find by taking any value of t > 0. So, if we let t = 1, say, we
have
c = F (1) = ln a − ln 1 = ln a,
as the value of the constant. Consequently, we have
F (t) = ln a = ln(at) − ln t,
which means that, if we rearrange this and let t = b where b > 0, we have
ln(ab) = ln a + ln b,
as required.
You can establish a slightly trickier identity for yourself in Activity 3.12.
Activity 3.12
Use integration to verify that
Z t
1
−1
dx.
tan t =
2
0 1+x
Hence show that if a is a constant and F is given by
a+t
−1
F (t) = tan
− tan−1 t,
1 − at
then F 0 (t) is zero. Deduce that F (t) is a constant and, by finding this constant,
deduce that
a+b
−1
−1
−1
tan a + tan b = tan
,
1 − ab
where b is a constant too.
81
3. The Riemann integral
Learning outcomes
At the end of this chapter and having completed the relevant reading and activities, you
should be able to:
use the definition of the Riemann integral to find the value of a definite integral;
3
use the Fundamental Theorem of Calculus to differentiate definite integrals;
use definite integrals and their derivatives in various applications.
Solutions to activities
Solution to activity 3.1
Given what we saw in Section 5.3 of 174 Calculus, we would expect the area, A, of the
region bounded by the curve y = 1 + x2 , the x-axis and the vertical lines x = −2 and
x = 2 to be given by the integral
Z
2
A=
(1 + x2 ) dx,
−2
as the integrand, 1 + x2 , is a non-negative function over the interval [−2, 2]. Evaluating
this integral, we then find that
x3
A= x+
3
2
−2
23
(−2)3
8
8
1
= 2+
− (−2) +
= 2+
− −2 −
=9 .
3
3
3
3
3
We can then use our results from Example 3.1, i.e. that L(P) = 5 and U (P) = 17, to
see that as
1
5 < 9 < 17,
3
they do indeed satisfy the inequality L(P) < A < U (P).
Solution to activity 3.2
Looking at the lower estimate, we have
1
0
1
1
1
2
1
n−1
L(Pn ) =
1+
+
1+
+
1+
+ ··· +
1+
n
n
n
n
n
n
n
n
!
1
0 + 1 + 2 + · · · + (n − 1)
1| + 1 + 1{z+ · · · + 1} +
=
n
n
n times
1
1 + 2 + · · · + (n − 1)
=
n+
n
n
1
1 n−1
=
n+
(n − 1 + 1) ,
n
n
2
82
3.2. Solutions to activities
where we have used the formula for the sum of the first n − 1 natural numbers to get
the term in the square brackets. If we now simplify this expression, we get
n−1
1 3n − 1
1
1
1
n+
=
=
3−
,
L(Pn ) =
n
2
n
2
2
n
as required. Similarly, looking at the upper estimate, we have
1
1
1
2
1
3
1
n
U (Pn ) =
1+
+
1+
+
1+
+ ··· +
1+
n
n
n
n
n
n
n
n
!
1
1 + 2 + 3 + ··· + n
1 + 1 + 1{z+ · · · + 1} +
=
|
n
n
n times
i
1
1 hn
=
n+
(n + 1) ,
n
n 2
3
where we have used the formula for the sum of the first n natural numbers to get the
term in the square brackets. If we now simplify this expression, we get
n+1
1 3n + 1
1
1
1
n+
=
=
3+
,
U (Pn ) =
n
2
n
2
2
n
as required.
Solution to activity 3.3
Looking at the illustrations in Figure 3.4, it should be clear that the area, A, of the
region bounded by the curve y = 1 + x, the x-axis and the vertical lines x = 0 and
x = 1 is 3/2 as it is the area of a trapezium (or the area of a right-angled triangle plus
the area of a square).10 We can then use our results from Example 3.2, i.e. that
1
1
1
1
L(Pn ) =
3−
and
U (Pn ) =
3+
,
2
n
2
n
to see that, as n > 0, we have
1
1
3− <3<3+
n
n
1
2
=⇒
1
3
1
1
3−
< <
3+
,
n
2
2
n
and so they do indeed satisfy the inequality L(Pn ) < A < U (Pn ).
10
Alternatively, we can see that it is given by the integral
Z 1
A=
(1 + x) dx,
0
as the integrand, 1 + x, is a non-negative function over the interval [0, 1]. Evaluating this integral, we
then find that
1 x2
12
02
1
3
A= x+
= 1+
− 0+
=1+ = ,
2 0
2
2
2
2
as we would expect from above.
83
3. The Riemann integral
Solution to activity 3.4
Suppose that f (x) is a non-negative continuous function defined over some interval [a, b]
and that P is the partition
{x0 , x1 , x2 , . . . , xn−1 , xn } where a = x0 < x1 < x2 < · · · < xn = b.
3
Now, if f (x) is an increasing function on the interval [a, b] and A is the area of the
region bounded by the curve y = f (x), the x-axis and the vertical lines x = a and x = b,
we can see that the
lower estimate is given by
L(P) =
n
X
i=1
(xi − xi−1 )mi =
n
X
i=1
(xi − xi−1 )f (xi−1 ),
where mi , the minimum value of f (x) for values of x in the sub-interval [xi−1 , xi ], is
simply f (xi−1 ) as the function f (x) is increasing over [a, b] and so it is increasing
over all of the sub-intervals as illustrated in Figure 3.10(a).
upper estimate is given by
U (P) =
n
X
i=1
(xi − xi−1 )Mi =
n
X
i=1
(xi − xi−1 )f (xi ),
where Mi , the maximum value of f (x) for values of x in the sub-interval [xi−1 , xi ],
is simply f (xi ) as the function f (x) is increasing over [a, b] and so it is increasing
over all of the sub-intervals as illustrated in Figure 3.10(b).
y
y
y = f (x)
y = f (x)
...
O x
0
x1
x2
...
(a)
...
xn−1 xn
x
O x
0
x1
x2
...
xn−1 xn
x
(b)
Figure 3.10: When f (x) is an increasing function. In (a) and (b) we have, respectively,
the rectangles that contribute towards the lower estimate, L(P), and the upper estimate,
U (P), of A based on the partition P. Note that, for clarity, only the first, second and last
rectangles are shown.
Of course, if f (x) is a decreasing function on the interval [a, b] and A is the area of the
region bounded by the curve y = f (x), the x-axis and the vertical lines x = a and x = b,
we can see that the
84
3.2. Solutions to activities
lower estimate is given by
L(P) =
n
X
i=1
(xi − xi−1 )mi =
n
X
i=1
(xi − xi−1 )f (xi ),
where mi , the minimum value of f (x) for values of x in the sub-interval [xi−1 , xi ], is
simply f (xi ) as the function f (x) is decreasing over [a, b] and so it is decreasing
over all of the sub-intervals.
upper estimate is given by
U (P) =
n
X
i=1
(xi − xi−1 )Mi =
n
X
i=1
(xi − xi−1 )f (xi−1 ),
where Mi , the maximum value of f (x) for values of x in the sub-interval [xi−1 , xi ],
is simply f (xi−1 ) as the function f (x) is decreasing over [a, b] and so it is decreasing
over all of the sub-intervals.
If you are having any trouble understanding this, try sketching the analogue of the
illustrations in Figure 3.10 with a function, f (x), which is now decreasing over the
interval [a, b].
Solution to activity 3.5
Comparing this with our answers from Example 3.1 and Activity 3.1, we now have the
values
1
5 < 6 < 9 < 14 < 17,
3
which means that we have
L(P) < L(P 0 ) < A < U (P 0 ) < U (P).
That is, both of our lower estimates are less than the value of A but L(P 0 ) is a better
lower estimate of A than L(P) whereas both of our upper estimates are greater than
the value of A but U (P 0 ) is a better upper estimate of A than U (P).
Indeed, it should be clear that this improvement in the lower and upper estimates has
come about because P 0 has one more point than P and this has enabled us to replace
the single rectangle for the sub-interval [−2, 0] from P (see Figure 3.3) with two
‘thinner’ rectangles for the sub-intervals [−2, −1] and [−1, 0] from P 0 (see Figure 3.5).
Solution to activity 3.6
If we sketch the curve y = 1 + x2 over the interval [0, 2] and indicate the points in the
partition Pm we see
k−1that
we will be looking at 2m rectangles whose bases are given by
k
the sub-intervals m , m for k = 1, 2, . . . , 2m. Indeed, we see that, for k = 1, 2, . . . , 2m,
each of these sub-intervals gives us a base whose length is
k
k−1
1
−
= ,
m
m
m
and so, for the
85
3
3. The Riemann integral
lower estimate, we need to sum the areas of the 2m rectangles illustrated in
Figure 3.11(a) where the height of each rectangle is given by the minimum value of
1 + x2 in each of the sub-intervals to get
1
L(Pm ) =
m
3
0
1+
m
2 !
1
+
m
1
1+
m
2 !
1
+ ··· +
m
2m − 1
1+
m
2 !
,
where we have taken each of the sub-intervals in turn and used ‘base times height’
to find the area of each rectangle. This gives us
!
02 + 12 + 22 + · · · + (2m − 1)2
1| + 1 + 1{z+ · · · + 1} +
m2
2m times
12 + 22 + · · · + (2m − 1)2
1
2m +
=
m
m2
1
1 2m − 1
=
2m + 2
([2m − 1] + 1)(2[2m − 1] + 1) ,
m
m
6
1
L(Pm ) =
m
where we have used the formula for the sum of the first 2m − 1 square numbers to
get the term in the big square brackets. If we now simplify this expression, we get
L(Pm ) =
=
=
=
∴ L(Pm ) =
1
(2m − 1)(2m)(4m − 1)
2m +
m
6m2
2m
2
6m + (2m − 1)(4m − 1)
6m3
1
2
2
6m + (8m − 6m + 1)
3m2
1
2
14m − 6m + 1
3m2
1
6m − 1
14 −
.
3
m2
Now we can see that, for m ≥ 1, we have
1
L(Pm ) =
3
6m − 1
14 −
m2
≤
14
,
3
and so L∗ = 14/3 is the smallest number that is larger than all the lower bounds.
upper estimate, we need to sum the areas of the 2m rectangles illustrated in
Figure 3.11(b) where the height of each rectangle is given by the maximum value of
1 + x2 in each of the sub-intervals to get
1
U (Pm ) =
m
1
1+
m
2 !
1
+
m
2
1+
m
2 !
1
+ ··· +
m
2m
1+
m
2 !
,
where we have taken each of the sub-intervals in turn and used ‘base times height’
86
3.2. Solutions to activities
to find the area of each rectangle. This gives us
2
2
1 + 2 + · · · + (2m)
m2
2m times
12 + 22 + · · · + (2m)2
1
2m +
=
m
m2
1
1 2m
=
2m + 2
(2m + 1)(2[2m] + 1) ,
m
m
6
U (Pm ) =
1
m
2
!
|1 + 1 + 1{z+ · · · + 1} +
3
where we have used the formula for the sum of the first 2m square numbers to get
the term in the big square brackets. If we now simplify this expression, we get
(2m)(2m + 1)(4m + 1)
1
2m +
U (Pm ) =
m
6m2
2m
2
=
6m + (2m + 1)(4m + 1)
6m3
1
2
2
=
6m + (8m + 6m + 1)
3m2
1
2
=
14m + 6m + 1
3m2
1
6m + 1
∴ U (Pm ) =
14 +
.
3
m2
Now we can see that, for m ≥ 1, we have
1
6m + 1
14
U (Pm ) =
14 +
≥ ,
2
3
m
3
and so U ∗ = 14/3 is the largest number that is smaller than all the upper bounds.
This means that there is only one number, I, that satisfies the inequality
L∗ ≤ I ≤ U ∗ ,
in this case, i.e. I = 14/3, and so we take this to be the value of the given Riemann
integral. That is,
Z 2
14
(1 + x2 ) dx = ,
3
0
and, indeed, this is what we should expect since
Z
0
2
2 x3
03
8
23
14
(1 + x ) dx = x +
= 2+
− 0+
=2+ = ,
3 0
3
3
3
3
2
if we evaluate the integral in the usual way.
87
3. The Riemann integral
y
y
y = 1 + x2
5
y = 1 + x2
5
3
...
...
1
O
1
1
m
2
m
...
2m−1
m
2
x
O
1
m
(a)
2
m
...
2m−1
m
2
x
(b)
Figure 3.11: In (a) and (b) we have, respectively, the rectangles that contribute towards
the lower estimates, L(Pm ), and the upper estimates, U (Pm ), based on the partition Pm .
Note that, for clarity, some of the y-intercepts have been omitted and only three of the
rectangles are shown.
Solution to activity 3.7
Looking at the lower estimates, we have
L(Pn ) =
=
=
=
0
1
1
1
2
1
n−2
1
1
2·
+
2·
+
2·
+ ··· +
2·
+
1
n
n
n
n
n
n
n
n
n
2 0 + 1 + 2 + · · · + (n − 2) 1
+
n
n
2
2 1 + 2 + · · · + (n − 2) 1
+
n
n
2
1
2 1 n−2
(n − 2 + 1) +
,
n n
2
2
where we have used the formula for the sum of the first n − 2 natural numbers to get
the term in the square brackets. If we now simplify this expression, we get
1
1
n−1
2
L(Pn ) = 2 (n − 2)(n − 1) + n = 2 n − 2n + 2 = 1 − 2 ·
,
n
n
n2
as required. Indeed, since n ≥ 2, we have n − 1 > 0 and n2 > 0, which means that
L(Pn ) = 1 − 2 ·
as we were asked to verify.
88
n−1
≤ 1,
n2
3.2. Solutions to activities
Similarly, for the upper estimates, we have
1
1
2
1
3
1
n−1
1 n
1
2·
+
2·
+
2·
+ ··· +
2·
+
2·
U (Pn ) =
n
n
n
n
n
n
n
n
n
n
2
= 2 1 + 2 + 3 + · · · + (n − 1) + n
n
2 n
= 2 (n + 1) ,
n 2
where we have used the formula for the sum of the first n natural numbers to get the
term in the square brackets. If we now simplify this expression, we get
U (Pn ) =
1
1
(n + 1) = 1 + ,
n
n
as required. Indeed, since n ≥ 2, we have
U (Pn ) = 1 +
1
≥ 1,
n
as we were asked to verify.
Solution to activity 3.8
As we are assuming that f is a non-negative function whose Riemann integral exists
over some interval containing the points a, b and c we know that all of the Riemann
integrals will represent the corresponding areas. So, by considering these areas, we can
see that Theorem 3.1(a) holds, i.e. that
Z a
f (x) dx = 0,
a
since the area bounded by the curve y = f (x), the x-axis and the vertical lines x = a
and x = a must be zero as the ‘base’ of this region is zero. We can also see that
Theorem 3.1(b) holds, i.e. that
Z b
Z c
Z b
f (x) dx =
f (x) dx +
f (x) dx,
a
a
c
since, for c ∈ [a, b], the area bounded by the curve y = f (x), the x-axis and the vertical
lines x = a and x = b is the same as the sum of the area bounded by the curve
y = f (x), the x-axis and the vertical lines x = a and x = c and the area bounded by the
curve y = f (x), the x-axis and the vertical lines x = c and x = b.
We can now deduce that Theorem 3.1(c) holds, i.e. that
Z b
Z a
f (x) dx = −
f (x) dx,
a
b
since, thinking about Theorem 3.1(a) and (b), we have
Z a
Z b
Z
0=
f (x) dx =
f (x) dx +
a
a
a
f (x) dx.
b
89
3
3. The Riemann integral
Of course, using this in Theorem 3.1(b), Theorem 3.1(d) follows immediately.
Solution to activity 3.9
Following on from what we saw in Section 3.2.1, when we have h < 0 so that t + h < t,
we are now interested in the difference between the two areas A(t + h) and A(t) given by
3
A(t) − A(t + h),
which is the area of the vertically hatched region of base |h| indicated in Figure 3.12. In
particular, we see that since f (x) is a decreasing function over the interval [a, b], we
have a
lower estimate of this area given by the area of the shaded rectangle in
Figure 3.9(a) which, using ‘base times height’, is
|h|f (t),
and so we have |h|f (t) ≤ A(t) − A(t + h).
upper estimate of this area given by the area of the shaded rectangle in
Figure 3.9(b) which, using ‘base times height’, is
|h|f (t + h),
and so we have A(t) − A(t + h) ≤ |h|f (t + h).
Consequently, putting these two inequalities together, we have
|h|f (t) ≤ A(t) − A(t + h) ≤ |h|f (t + h)
=⇒
f (t) ≤
A(t) − A(t + h)
≤ f (t + h),
|h|
as |h| > 0. Indeed, as h < 0, we have |h| = −h and so this gives us
f (t) ≤
A(t) − A(t + h)
≤ f (t + h)
−h
=⇒
f (t) ≤
A(t + h) − A(t)
≤ f (t + h),
h
which means that, taking the limit as h → 0− on both sides of this inequality, we get
lim− f (t) ≤ lim−
h→0
h→0
A(t + h) − A(t)
≤ lim− f (t+h)
h→0
h
=⇒
f (t) ≤ lim−
h→0
A(t + h) − A(t)
≤ f (t),
h
if we use the continuity of f (x) over [a, b].11 Of course, this means that
lim−
h→0
A(t + h) − A(t)
= f (t),
h
as required.
11
Notice that the continuity of f (x) over [a, b] is essential here in order to justify the assertion that
the limit
lim− f (t + h),
h→0
exists and that it is equal to f (t). Also, notice that we can only take the limit as h → 0− here as we
have assumed that h < 0.
90
3.2. Solutions to activities
y
O
y
a
c
0000
011111
0000
101111
101111
0000
101111
0000
101111
0000
101111
0000
10
1111
0000
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
y = f (x)
t+h t
x
b
O
a
c
t+h t
(a)
3
y = f (x)
b
x
(b)
Figure 3.12: The areas of the shaded regions in (a) and (b) are the lower and upper
estimates, respectively, of the area A(t) − A(t + h) of the vertically hatched region. Notice
that here we have h < 0 and so the ‘base’ of the rectangles is |h|.
Solution to activity 3.10
If, as instructed, we write F (x) as a definite integral of the form
Z x
F (x) =
f (t) dt,
A
then, assuming that A is a constant,12 the FTC tells us that
Z x
d
0
f (t) dt = f (x),
F (x) =
dx
A
2
and, as we are told that F 0 (x) = e−x , this gives us
−x2
f (x) = e
Z
=⇒
x
F (x) =
2
e−t dt,
A
for a suitable choice of the [assumed] constant, A. Now, we are also told that F (0) = 0,
which means that
Z 0
2
F (0) =
e−t dt = 0,
A
2
and, since this is the area between the positive function e−t , the t-axis and the vertical
lines t = A and t = 0, this means that we must take A = 0 [which is a constant] for this
to hold. Consequently, we see that
Z x
2
F (x) =
e−t dt,
0
is the required definite integral.
12
If this assumption turns out to give us nonsense, then we would have to try something more
complicated.
91
3. The Riemann integral
Solution to activity 3.11
Here, we have
Z
t3
F (t) =
4t
3
x3
x dx =
3
2
t3
=
4t
t9 64
(t3 )3 (4t)3
−
= − t3 ,
3
3
3
3
and so
F 0 (t) = 3t8 − 64t2 ,
in agreement with what we found in Example 3.11.
Solution to activity 3.12
Firstly, it is easy to verify that we have
t
Z t
1
−1
dx = tan x = tan−1 t − tan−1 0 = tan−1 t,
2
0 1+x
0
using the result from Example 5.17 of 174 Calculus and noting that tan−1 0 = 0 because
tan 0 = 0. Then, if a is a constant, we can use this to write
F (t) = tan
−1
a+t
1 − at
−1
− tan
Z
t=
a+t
1−at
0
1
dx −
1 + x2
Z
0
t
1
dx,
1 + x2
so that, using the FTC and its extension as appropriate, we have
!
(1)(1
−
at)
−
(a
+
t)(−a)
1
1
,
F 0 (t) =
−
2
(1 − at)2
1 + t2
1 + a+t
1−at
where we have used the quotient rule to differentiate the upper limit of the first
integral. Now, considering the first term, we have
!
(1)(1 − at) − (a + t)(−a)
1
1 − at + a2 + at
=
2
(1 − at)2
(1 − at)2 + (a + t)2
1 + a+t
1−at
1 + a2
(1 − 2at + a2 t2 ) + (a2 + 2at + t2 )
1 + a2
=
1 + a2 t2 + a2 + t2
1 + a2
=
(1 + a2 )(1 + t2 )
1
=
,
1 + t2
=
which means that
1
1
−
= 0,
2
1+t
1 + t2
as required. But, of course, if F 0 (t) = 0, this means that F (t) = c where c is a constant
which we can find by taking any value of t. So, if we let t = 0, say, we have
F 0 (t) =
c = F (0) = tan−1 a − tan−1 0 = tan−1 a,
92
3.2. Exercises
as the value of the constant. Consequently, we have
a+t
−1
−1
F (t) = tan a = tan
− tan−1 t,
1 − at
which means that, if we rearrange this and let t = b, we have
a+b
−1
−1
−1
,
tan a + tan b = tan
1 − ab
3
as required.
Exercises
Exercise 3.1
Consider, for n ∈ N, the partition P = {1, 2, . . . , n} of the interval [1, n]. Find the lower
and upper estimates, L(P) and U (P) respectively, of the integral
Z n
1
dx,
1 x(x + 1)
simplifying your answers as far as possible.
Exercise 3.2
Consider, for n ∈ N, the partition
Pn = 0, n1 , n2 , . . . , 1 ,
of the interval [0, 1]. Find the lower and upper estimates, L(Pn ) and U (Pn ) respectively,
of the integral
Z 1
ex dx,
0
simplifying your answers as far as possible.
Find lim L(Pn ) and lim U (Pn ). What do these tell us about the value of the integral?
n→∞
n→∞
Exercise 3.3
Z
0
sin t
Find f (t) when f (t) =
x3 dx.
cos t
Exercise 3.4
Find the following limits.
(a)
lim+ t ln t,
t→0
Z
Hence find lim+ (ln t)
t→0
(b)
t
lim+ t ,
t→0
(c)
1
lim+
t→0 t
Z
t
xx dx.
0
t
xx dx.
0
93
3. The Riemann integral
Exercise 3.5
Let f be a continuous function that takes positive real values and suppose that
Z t2
G(t) =
f (x) dx.
t
3
0
Find G (t) and hence use a Taylor series to find a first-order approximation to G(t) for
values of t close to 1.
Solutions to exercises
Solution to exercise 3.1
If we sketch the curve y = 1/x(x + 1) over the interval [1, n] and indicate the points in
the partition P we see that we will be looking at n − 1 rectangles whose bases are given
by the sub-intervals [k − 1, k] for k = 2, 3, . . . , n. Indeed, we see that, for k = 2, 3, . . . , n,
each of these sub-intervals gives us a base whose length is
k − (k − 1) = 1,
and so, for the
lower estimate, we need to sum the areas of the n − 1 rectangles illustrated in
Figure 3.13(a) where the height of each rectangle is given by the minimum value of
1/x(x + 1) in each of the sub-intervals to get
1
1
1
1
+1
+1
+ ··· + 1
L(P) = 1
2(2 + 1)
3(3 + 1)
4(4 + 1)
n(n + 1)
=
1
1
1
1
+
+
+ ··· +
,
2·3 3·4 4·5
n(n + 1)
where we have taken each of the sub-intervals in turn and used ‘base times height’
to find the area of each rectangle. Then, using partial fractions, we see that
1
1
1
= −
,
k(k + 1)
k k+1
and so we can write
1 1
1 1
1 1
1
1
1
1
L(P) =
−
+
−
+
−
+ ··· +
−
= −
,
2 3
3 4
4 5
n n+1
2 n+1
as the intermediate terms cancel.
upper estimate, we need to sum the areas of the n − 1 rectangles illustrated in
Figure 3.13(b) where the height of each rectangle is given by the maximum value of
1/x(x + 1) in each of the sub-intervals to get
1
1
1
1
U (P) = 1
+1
+1
+ ··· + 1
1(1 + 1)
2(2 + 1)
3(3 + 1)
(n − 1)(n − 1 + 1)
=
94
1
1
1
1
+
+
+ ··· +
,
1·2 2·3 3·4
(n − 1)n
3.2. Solutions to exercises
where we have taken each of the sub-intervals in turn and used ‘base times height’
to find the area of each rectangle. Then, using partial fractions again, we can write
1 1
1 1
1
1
1
1 1
−
+
−
+
−
+ ··· +
−
=1− ,
U (P) =
1 2
2 3
3 4
n−1 n
n
as the intermediate terms cancel.
3
Thus, we have found that the lower and upper estimates of the given integral are
L(P) =
1
1
−
2 n+1
U (P) = 1 −
and
1
,
n
respectively.
y
y
1
x(x + 1)
y=
...
O
1 2 3 4
...
(a)
1
x(x + 1)
y=
...
n−1 n
x
O
...
1 2 3 4
(b)
n−1 n
x
Figure 3.13: The sketches for Exercise 3.1. In (a) and (b) we have, respectively, the
rectangles that contribute towards the lower estimates, L(P), and the upper estimates,
U (P), based on the partition P. Note that, for clarity, the y-intercepts have been omitted
and only four of the rectangles are shown.
Solution to exercise 3.2
If we sketch the curve y = ex over the interval [0, 1] and indicate the points in the
partition Pn we
we will be looking at n rectangles whose bases are given by the
seekthat
,
for
k = 1, 2, . . . , n. Indeed, we see that, for k = 1, 2, . . . , n, each of
sub-intervals k−1
n
n
these sub-intervals give us a base whose length is
k k−1
1
−
= ,
n
n
n
and so, for the
lower estimate, we need to sum the areas of the n rectangles illustrated in
Figure 3.14(a) where the height of each rectangle is given by the minimum value of
ex in each of the sub-intervals to get
L(Pn ) =
1 0 1 1/n
1
e + e + · · · + e(n−1)/n ,
n
n
n
95
3. The Riemann integral
where we have taken each of the sub-intervals in turn and used ‘base times height’
to find the area of each rectangle. This gives us
1 0
e + e1/n + · · · + e(n−1)/n
n
1
1 + e1/n + · · · + e(n−1)/n
=
n
1 1 − (e1/n )n
=
n
1 − e1/n
1
1−e
=
,
n 1 − e1/n
L(Pn ) =
3
where we have used the formula for the sum of a geometric series to get the final
answer. We now see that writing
1−e
1
1/n
,
lim L(Pn ) = lim
= (e −1) lim 1/n
1/n
n→∞
n→∞ n
n→∞ e
1−e
−1
we are dealing with the limit of a quotient where the numerator and the
denominator both tend to zero as n → ∞. As such, we can use L’Hôpital’s rule to
see that
1
−1/n2
= (e −1) lim 1/n = e −1,
2
1/n
n→∞ e
n→∞ (−1/n ) e
lim L(Pn ) = (e −1) lim
n→∞
is the first of the sought after limits.
upper estimate, we need to sum the areas of the n rectangles illustrated in
Figure 3.14(b) where the height of each rectangle is given by the maximum value of
ex in each of the sub-intervals to get
U (Pn ) =
1
1 1/n 1 2/n
e + e + · · · + e1 ,
n
n
n
where we have taken each of the sub-intervals in turn and used ‘base times height’
to find the area of each rectangle. This gives us
1 1/n
e + e2/n + · · · + e1
n
e1/n
=
1 + e1/n · · · + e(n−1)/n
n 1/n
e
1 − (e1/n )n
=
n
1 − e1/n
e1/n
1−e
=
,
n
1 − e1/n
U (Pn ) =
where we have used the formula for the sum of a geometric series to get the final
answer. We now see that writing
e1/n
1−e
1/n
lim U (Pn ) = lim
= (e −1) lim
,
1/n
n→∞
n→∞ n
n→∞
1−e
1 − e−1/n
96
3.2. Solutions to exercises
we are dealing with the limit of a quotient where the numerator and the
denominator both tend to zero as n → ∞. As such, we can use L’Hôpital’s rule to
see that
−1/n2
1
=
(e
−1)
lim
= e −1,
n→∞ −(1/n2 ) e−1/n
n→∞ e−1/n
lim U (Pn ) = (e −1) lim
n→∞
3
is the second of the sought after limits.
What do these limits tell us about the value of the integral? Well, as n increases, the
estimates we get from the partition Pn should get better and better so that, in
particular, as we take the limit as n → ∞, we get the best estimates that Pn can provide.
Indeed, as these best lower and upper estimates are the same, this should lead us to
conclude that this common value is the value of the integral, i.e. we have found that
Z 1
ex dx = e −1,
0
which is, of course, exactly what we would expect to get if we found the integral in the
usual way!
y
y
y = ex
y = ex
...
...
1
O
1
1
n
2
n
...
n−1
n
1
x
O
(a)
1
n
2
n
...
n−1
n
1
x
(b)
Figure 3.14: The sketches for Exercise 3.2. In (a) and (b) we have, respectively, the
rectangles that contribute towards the lower estimates, L(Pn ), and the upper estimates,
U (Pn ), based on the partition Pn . Note that, for clarity, the y-intercepts have been
omitted and only three of the rectangles are shown.
Solution to exercise 3.3
Given that
Z
sin t
f (t) =
x3 dx,
cos t
we can apply our extension to the FTC from Section 3.2.5 to see that
f 0 (t) = (cos t)(sin t)3 − (− sin t)(cos t)3 = sin t cos t(sin2 t + cos2 t) = sin t cos t,
97
3. The Riemann integral
if we use the trigonometric identity sin2 t + cos2 t = 1.
Solution to exercise 3.4
For (a), we write
ln t
,
t→0
t→0 1/t
so that we have a quotient where the numerator and the denominator both tend to
infinity [in magnitude] as t → 0+ . As such, we can use L’Hôpital’s rule to see that
lim+ t ln t = lim+
3
lim+ t ln t = lim+
t→0
t→0
1/t
ln t
= lim+
= lim −t = 0.
1/t t→0 −1/t2 t→0+
For (b), we use the fact that tt = et ln t so that
lim tt = lim+ et ln t = e0 = 1,
t→0+
t→0
if we use our answer from (a).
For (c), we write
Rt x
Z
x dx
1 t x
,
lim+
x dx = lim+ 0
t→0 t 0
t→0
t
so that we have a quotient where the numerator and the denominator both tend to zero
as t → 0+ . As such, we can use L’Hôpital’s rule and the FTC to see that
Rt x
Z
x dx
1 t x
tt
lim+
= lim+ = 1,
x dx = lim+ 0
t→0
t→0 t 0
t→0 1
t
if we use our answer from (b).
Then, lastly, we can see that if we use our results from (a) and (c), we have
Z t
Z t
1
x
x
lim (ln t)
x dx = lim+ t ln t
x dx = (0)(1) = 0,
t→0+
t→0
t 0
0
as the final answer.
Solution to exercise 3.5
If f is a continuous function that takes positive real values and we are given that
Z t2
G(t) =
f (x) dx,
t
then, using our extension of the FTC from Section 3.2.5, we find that
G0 (t) = (2t)f (t2 ) − (1)f (t) = 2tf (t2 ) − f (t).
As such, noting that
Z 1
G(1) =
f (x) dx = 0
1
and
G0 (1) = 2(1)f (1) − f (1) = f (1),
we see that a first-order Taylor series about t = 1 gives us
G(t) = G(1) + (t − 1)G0 (1) + · · · = 0 + (t − 1)f (1) + · · · = (t − 1)f (1) + · · · ,
and so (t − 1)f (1) is the required first-order approximation to G(t) for values of t close
to 1.
98