Limits and Continuity
Philippe B. Laval
Kennesaw State University
January 12, 2005
Abstract
Notes and practice problems on limits and continuity.
Contents
1 Limits
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Theory: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2.1 Graphical Method . . . . . . . . . . . . . . . . . . . . . .
1.2.2 Numerical Method . . . . . . . . . . . . . . . . . . . . . .
1.2.3 Analytical Method . . . . . . . . . . . . . . . . . . . . . .
1.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3.1 Numerical Method . . . . . . . . . . . . . . . . . . . . . .
1.3.2 Direct application of the limit rules or of one of the two
theorems above . . . . . . . . . . . . . . . . . . . . . . . .
1.3.3 Rational function for which both the numerator and denominator approach 0 . . . . . . . . . . . . . . . . . . . .
1.3.4 Fraction whose denominator and numerator go to 0 and
whose numerator or denominator contains a radical . . .
1.3.5 Function whose behavior changes depending on whether
x > a or x < a . . . . . . . . . . . . . . . . . . . . . . . .
1.3.6 Squeeze Theorem . . . . . . . . . . . . . . . . . . . . . . .
1.4 Limits Involving Infinity . . . . . . . . . . . . . . . . . . . . . . .
1.4.1 If f gets arbitrarily large . . . . . . . . . . . . . . . . . .
1.4.2 Limits at infinity that is limits when x approaches −∞ or
∞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5 Sample problems: . . . . . . . . . . . . . . . . . . . . . . . . . .
2
2
2
3
3
3
4
4
5
5
6
7
8
8
8
10
11
2 Continuity
13
2.1 Theory: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.3 Sample problems: . . . . . . . . . . . . . . . . . . . . . . . . . . 16
1
3 Limit Laws
1
17
Limits
1.1
Introduction
We are trying to study the behavior of a given function f (x) near a given value
x = a. This is denoted by the symbol lim f (x). It is read "limit of f (x) as
x→a
x approaches a". As you know, given a function y = f (x), to each value of
x corresponds a unique value y = f (x). As we plug in various values of x,
we obtain corresponding values for y. We are interested in knowing that if the
values of x follows a certain pattern, do the corresponding values of y follow a
similar pattern? More specifically, the pattern we are interested in is values of
x such that x approaches a given value a. This is denoted by x → a. As we will
see, there are several possibilities. If y = f (x) and x → a, then we could have:
1. The corresponding values of y also approach some number we will call L.
In this case, we will write lim f (x) = L. We also say y → L as x → a (y
x→a
approaches L as x approaches a).
2. The corresponding values of y get arbitrarily large (go to infinity) or arbitrarily small (go to minus infinity). In this case, we write lim f (x) = ∞
x→a
or lim f (x) = −∞.
x→a
3. The corresponding values of y do not seem to follow any pattern.
In this section, we will learn how to determine which situation we are in.
Since a function can be given different ways (graph, table, formula), we must
learn how to do this in all these cases.
1.2
Theory:
Definition 1 (Limit of a function at a point) lim f (x) means that we are
x→a
looking at the values of f (x) for values of x close to a but not equal to a. In fact,
f may not even be defined at a. If we can make f (x) as close as we want to a
number L simply by taking x close enough to a, then we say that lim f (x) = L.
x→a
Remark 2 When we say ”x close to a”, we must realize that x can get closer
to a from the right (i.e. x is getting closer to a but is larger than a). But, x can
also get closer to a from the left (i.e. x is getting closer to a but is smaller than
a). Thus, we have the two additional limits, which are called one-sided limits
Definition 3 (One-sided limits) When we say ”x approaches a”, we need to
realize that x can get close to a from the right as well as from the left. If we
wish to consider only one side, we have the following one-sided limits:
2
1. lim+ f (x) means that we are looking at values of f (x) for x close to a
x→a
and larger than a (to the right of a).
2. lim− f (x) means that we are looking at values of f (x) for x close to a
x→a
and smaller than a (to the left of a).
One condition for lim f (x) to exist is that both lim+ f (x) and lim− f (x)
x→a
x→a
x→a
exist and are equal.
There are three methods used to evaluate the limit of a function at a point:
1.2.1
Graphical Method
In this case, the graph of the function is used to estimate the values of f (x) as
x gets closer and closer to a. See your book on pages 102, 103 for nice pictures.
1.2.2
Numerical Method
In this case, we use the formula defining the function to compute the values of
f (x) as x gets closer and closer to a. When I want students to use this method,
I will ask the students to ”guess” the limit. It is a good idea to use two tables.
One table will have the values of x close to a but less than a. This will allow
us to guess lim f (x). The other will have the values of x close to a but larger
x→a−
than a. This will allow us to guess lim+ f (x). Consider the following example:
x→a
Example 4 Guess lim x3
x→2
We construct the two tables mentioned above.
x
1.9
1.99
1.999 1.9999
and
x3 6.859 7.8806 7.988 7.9988
x
2.1
2.01
2.001 2.0001
3
x
9.261 8.1206 8.012 8.0012
From the table, we see that as x gets closer to 2, from either the right or the
left, x3 is getting closer to 8. We conclude that lim x3 = 8
x→2
However, you should use caution with this method. See examples 4 and 5
on pages 104, 105.
1.2.3
Analytical Method
In this case, the limit rules are used. (See a list of the limit rules at the end of
this document). Using the limit rules, the following two theorems can be proven
Theorem 5 (Limit of a polynomial) If f (x) is a polynomial, then lim f (x) =
x→a
f (a). In other words, we just plug in the value of a in the polynomial.
3
Theorem 6 (Limit of a rational function) If f (x) is a rational function
p (x)
that is if there exists two polynomials p (x) and q (x) such that f (x) =
q (x)
then
lim f (x) = f (a) providing lim q (x) = 0
x→a
x→a
In other words, we just plug in the value of a in the rational function as long as
q (a) = 0.
1.3
Examples
We look at several examples and how to handle them. The way limits are
evaluated depends a lot on the type of function involved. The examples below
are classified into various categories. You should first establish which category
the function falls under, then follow the steps described for that category. The
cases below assume we are trying to evaluate lim f (x).
x→a
1.3.1
Numerical Method
Students are often confused regarding the values of x to try and how many to
try. There is not a definite rule. The key is to try values which follow the given
pattern. Enough velues should be tried to allow seeing a pattern if there is one.
Let us consider a specific example. Suppose that we are computing the limit
of some function as x → 2. For values of x approaching 2 from the right, one
could try 2.1, 2.01, 2.001, 2.0001. These values are clearly getting closer and
closer to 2. The values should get very close to 2, which these do. 3 or 4 values
should be tried. If there is a pattern in the corresponding y − values, 4 values
will allow to see it. For values of x approaching 2 from the left, one could try
1.9, 1.99, 1.999, 1.9999.
Let us now look at specific examples.
x2 − 1
x→1 x − 1
We are supposed to try values of x which get closer and closer to 1 from both
the left and the right. To do so, we build the following tables
x
1.1 1.01 1.001 1.0001
and
x2 − 1
2.1 2.01 2.001 2.0001
x−1
Example 7 Guess lim
x
x2 − 1
x−1
.9
.99
.999
.9999
1.9
1.99
1.999
1.9999
Thus, it appears that as x → 1,
x2 − 1
= 2.
x→1 x − 1
x2 − 1
is getting closer and closer to 2. We
x−1
write lim
4
Example 8 Guess lim sin πx
x→0
We are supposed to try values of x which get closer and closer to 0 from both
the left and the right. To do so, we build the following tables
x .1 .01 .001 .0001
and
sin πx
0 0
0
0
x -.1 -.01 -.001 -.0001
sin πx
0
0
0
0
So, we might be tempted to conclude that lim sin πx = 0. This is not correct
x→0
however. This limit is not defined. See example 4 on page 104. A good way to
avoid this is to break the pattern in the values of x we
For example, if we
try.
π had tried x = .0015 then we would have seen that sin
= 0.866 03 which
.0015
is not zero.
1.3.2
Direct application of the limit rules or of one of the two theorems above
Example 9 Find lim x3 − 5x + 12
x→0
This is a polynomial, to find the limit, we simply plug in the point. Thus,
lim x3 − 5x + 12 = 12
x→0
x2 + 1
x→2 x − 5
This is a rational function, its denominator is not zero when x = 2, thus, by
the theorem on ”limit of a rational function”, we evaluate the limit by plugging
in the point. Thus,
Example 10 Find lim
x2 + 1
22 + 1
=
x→2 x − 5
2−5
5
=
−3
5
=−
3
lim
Example 11 Find lim
√
2x2 − 1
This looks like rule # 11. Since lim 2x2 − 1 = 17 > 0, it follows that
x→3
x→3
lim
x→3
1.3.3
√
2x2 − 1 = 17
Rational function for which both the numerator and denominator approach 0
In this case, none of the rules work. The theorem on rational function does
not work either since the denominator goes to 0. Since both the numerator
5
and denominator go to 0 as x approaches a, we can factor x − a from both the
numerator and denominator.
x2 − 1
x→1 x − 1
This is indeed a rational function, both the numerator and denominator are 0
when x is 1. We proceed as follows:
Example 12 Find lim
x2 − 1
(x − 1) (x + 1)
= lim
(factor the numerator)
x→1 x − 1
x→1
x−1
= lim (x + 1)
lim
x→1
= 2 (since x + 1 is a polynomial)
x2 + x − 6
x→2 x2 − x − 2
Example 13 Find lim
x2 + x − 6
(x − 2) (x + 3)
= lim
2
x→2 x − x − 2
x→2 (x − 2) (x + 1)
(x + 3)
= lim
x→2 (x + 1)
5
=
3
lim
1.3.4
Fraction whose denominator and numerator go to 0 and whose
numerator or denominator contains a radical
In this case, try to rationalize the part which contains the radical. This means
multiplying by the conjugate. Recall, the conjugate of an expression of the form
a + b is a − b. We will use the identity (a + b) (a − b) = a2 − b2 .
√
x−2
Example 14 Find lim
x→4 x − 4
√
√
√
x−2
x−2 x+2
√
= lim
lim
x→4 x − 4
x→4 x − 4
x+2
√
√
( x − 2) ( x + 2)
√
= lim
x→4 (x − 4) ( x + 2)
√ 2
( x) − 4
√
= lim
x→4 (x − 4) ( x + 2)
x−4
√
= lim
x→4 (x − 4) ( x + 2)
1
= lim √
x→4
x+2
1
=
4
6
1.3.5
Function whose behavior changes depending on whether x > a
or x < a
In this case, the one sided limits lim− f (x) and lim+ f (x) must be evaluated.
x→a
x→a
This is the case when we have a piecewise function and we are evaluating the
limit at the breaking point. It is also the case if we have absolute values and
the point at which we are evaluating the limit is the point where the expression
inside the absolute values changes sign. The examples below illustrate this:
|x − 2|
x−2
We note that x − 2 changes sign at 2, the point where we are evaluating the
limit. So, instead, we compute
Example 15 Find lim
x→2
|x − 2|
x−2
= lim
lim+
x→2
x−2
x−2
x→2
(if x → 2+ then x > 2, so x − 2 > 0 hence |x − 2| = x − 2)
= lim (1)
x→2
=1
and also
lim
x→2−
|x − 2|
− (x − 2)
= lim
x→2
x−2
x−2
= lim (−1)
x→2
= −1
Since the one-sided limits are not equal, we conclude that lim
x→2
exist.
|x − 2|
does not
x−2
Remark 16 If we had computed the limit at any number other than 2, there
would not have been a problem because in that case, x−2 would not have changed
sign.
x + 1 if x > 2
Example 17 Find lim f (x) where f (x) =
x − 2 if x < 2
x→2
Once again, the fact that we want the limit as x approaches 2 is crucial. 2 is
the point where the definition of f changes. So, we have to compute
lim f (x) = lim+ (x + 1)
x→2
x→2+
=3
and
lim f (x) = lim− (x − 2)
x→2−
x→2
=0
7
Since the one-sided limits do not agree, we conclude that lim f (x) does not
x→2
exist.
1.3.6
Squeeze Theorem
Theorem 18 If f (x) ≤ g(x) ≤ h(x) for all x in an open interval containing
a, except possibly at a, and if lim f (x) = lim h(x) = L then lim g(x) = L
x→a
x→a
x→a
This is used as follows. Supposed we are trying to evaluate the limit of some
function g, i.e. we want to find lim g(x). The function g is such that it does
x→a
not fit any of the cases described above. We may try to find two functions f
and h which satisfy the conditions of the theorem. If we do, then we will know
lim g(x).
x→a
1
Example 19 Find lim x sin
x→0
x
2
Our goal is to find two functions, one larger, one smaller than x2 sin
the same limit as x approaches 0. We start by noticing that
1
≤1
−1 ≤ sin
x
1
−x2 ≤ x2 sin
≤ x2 (since x2 is positive)
x
1
having
x
Clearly, lim x2 = lim −x2 = 0. By the squeeze theorem, lim x2 sin
0
1.4
x→0
x→0
x→0
1
=
x
Limits Involving Infinity
There are two cases to consider. First, the function could go to infinity (i.e.
get arbitrarily large) when x approaches a finite number. Second, x could go to
infinity.
1.4.1
If f gets arbitrarily large
This happens when we have a fraction whose numerator approaches a finite
number and whose denominator approaches 0. In this case, you will have to
look at the one-sided limits. These one-sided limits will be ±∞, the sign being
determined by the sign of the fraction. Also, remember that ln x approaches
−∞ as x → 0+ therefore, lim+ ln x = −∞.
x→0
1
Example 20 Find lim 2
x→0 x
The numerator of this fraction is non zero while the denominator approaches 0.
8
Thus, we look at
x→0+
lim
1
=∞
x2
lim−
1
=∞
x2
lim
1
=∞
x2
and
x→0
Thus
x→0
1
x
The numerator of this fraction is non zero while the denominator approaches 0.
Thus, we look at
1
=∞
lim
x→0+ x
and
1
lim− = −∞
x→0 x
Thus
1
lim does not exist
x→0 x
1−x
Example 22 Find lim
x→2 x − 2
The numerator of this fraction approaches a non-zero number while the denominator approaches 0. Thus, we look at
Example 21 Find lim
x→0
lim
x→2+
1−x
= −∞
x−2
and
lim
x→2−
Thus
lim
x→2
1−x
=∞
x−2
1−x
does not exist
x−2
Definition 23 (Vertical asymptote) We say that the line x = a is a vertical
asymptote of the curve y = f (x) if one of the conditions below is satisfied:
1. lim f (x) = ±∞
x→a
2. lim+ f (x) = ±∞
x→a
3. lim− f (x) = ±∞
x→a
Remark 24 The function ln x has a vertical asymptote at x = 0. The function
π
3π
5π
tan x has vertical asymptotes at ± , ± , ± , ...
2
2
2
See your book on page 132 for pictures of asymptotes.
9
1.4.2
Limits at infinity that is limits when x approaches −∞ or ∞
In this case, the limit could be finite, infinite or may not exist.
Definition 25 (Horizontal asymptote) We say that the line y = a is a horizontal asymptote of the curve y = f (x) if one of the conditions below is satisfied:
1. lim f (x) = a
x→∞
2.
lim f (x) = a
x→−∞
Theorem 26 The following limits at infinity are important:
1.
2.
3.
lim
x→±∞
1
= 0 if n is a positive integer.
xn
lim xn = ∞ if n is an even positive integer.
x→±∞
lim xn = ±∞ if n is an odd positive integer.
x→±∞
4. lim ex = ∞
x→∞
5.
lim ex = 0
x→−∞
6. lim ln x = ∞
x→∞
7. The limits of the trigonometric functions do not exist at infinity.
Remark 27 When finding the limit at infinity of a polynomial or a rational
function, factor the term of highest degree. See examples below.
Remark 28 When evaluating limits, if you find one of the forms below, you
won’t be able to conclude. You will have to try to simplify further. These forms
are called indeterminate forms because their behavior is not predictable.
1. ∞ − ∞
∞
2.
∞
0
3.
0
4. 0 · ∞
Example 29 Find lim x2 − x
x→∞
If we try to simply evaluate, we get ∞ − ∞. Remember, this is not 0. The
correct way to do it is as follows:
1
lim x2 − x = lim x2 1 −
x→∞
x→∞
x
= lim x2
x→∞
=∞
10
x2 + 5x − 3
x→∞ 3x2 + 2x
Example 30 Find lim
5
3
2
x
1
+
−
x2 + 5x − 3
x x2
lim
=
lim
x→∞ 3x2 + 2x
x→∞
2
2
3x 1 +
3x
x2
= lim
x→∞ 3x2
1
=
3
Example 31 Find the vertical and horizontal asymptotes of f (x) =
x+1
x−1
• Vertical asymptotes.
We know these happen if f goes to infinity when x approaches a finite
number. Since f is a fraction, it will go to infinity if its denominator goes
to 0 and not its numerator. It is easy to see that the denominator of f
goes to 0 when x approaches 1. The line x = 1 is a vertical asymptote.
• Horizontal asymptotes.
These are found by computing lim f (x) and lim f (x)
x→∞
x→−∞
x+1
x−1
1
x 1+
x
= lim x→∞
1
x 1−
x
=1
lim f (x) = lim
x→∞
x→∞
Similarly,
x+1
x−1
1
x 1+
x
= lim
x→−∞
1
x 1−
x
=1
lim f (x) = lim
x→−∞
x→−∞
So, the line y = 1 is a horizontal asymptote.
1.5
Sample problems:
1. Do # 3, 5, 7, 13 on pages 108, 109
11
2. Do # 1, 2, 3, 7, 9, 11, 15, 24 on pages 117, 118
x2 + 1
(use table of values) (answer: does not exist)
x→0
x
3
4. Guess lim+
(use table of values) (answer: ∞)
x→2 x − 2
3. Guess lim
x2 + 2x − 8
(answer: 8.0)
x→4
x−2
5. Evaluate lim
x2 + 2x − 8
(answer: 6.0)
x→2
x−2
√
x − 3x − 2
(answer: 0.0)
7. Evaluate lim
x→1
x2 − 4
√
x − 3x − 2
1
answer:
8. Evaluate lim
x→2
x2 − 4
16
6. Evaluate lim
9. Evaluate lim
x→3
10. Evaluate lim
x→1
11. Let f (x) =
⎧
⎨
|x − 1|
(answer: 1.0)
x−1
|x − 1|
(answer: does not exist)
x−1
x
if
x<0
if 0 < x ≤ 2
x2
⎩
8 − x if
x>2
(a) Evaluate lim f (x) (answer 1.0)
x→1
(b) Evaluate lim− f (x) (answer: 0.0)
x→0
(c) Evaluate lim+ f (x) (answer: 0.0)
x→0
(d) Evaluate lim f (x) (answer: 0.0)
x→0
(e) Evaluate lim− f (x) (answer: 4.0)
x→2
(f) Evaluate lim+ f (x) (answer: 6.0)
x→2
(g) Evaluate lim f (x) (answer: does not exist)
x→2
x − x2
(answer: 0.0)
x→∞ x4 − 5
12. Find lim
3
4x3 − x2
(answer: 4.0)
x→∞ x3 − 5x + 1
13. Find lim
1
(answer: horizontal asymptotes:
−1
y = 0, vertical asymptotes: x = −1 and x = 1)
14. Find the asymptotes of f (x) =
x2
12
2
Continuity
2.1
Theory:
Definition 32 (Continuity) A function f is said to be continuous at x = a
if the three conditions below are satisfied:
1. a is in the domain of f (i.e. f (a) exists )
2. lim f (x) exists
x→a
3. lim f (x) = f (a)
x→a
Definition 33 (Continuity from the left) A function f is said to be continuous from the left at x = a if the three conditions below are satisfied:
1. a is in the domain of f (i.e. f (a) exists )
2. lim− f (x) exists
x→a
3. lim f (x) = f (a)
x→a−
Definition 34 (Continuity from the right) A function f is said to be continuous from the right at x = a if the three conditions below are satisfied:
1. a is in the domain of f (i.e. f (a) exists )
2. lim+ f (x) exists
x→a
3. lim+ f (x) = f (a)
x→a
Definition 35 (Continuity on an interval) A function f is said to be continuous on an interval I if f is continuous at every point of the interval.
If a function is not continuous at a point x = a, we say that f is discontinuous
at x = a. When looking at the graph of a function, one can tell if the function
is continuous because the graph will have no breaks or holes. One should be
able to draw such a graph without lifting the pen from the paper.
In general, functions are continuous almost everywhere. The only places to
watch are places where the function will not be defined, or places where the
definition of the function changes (breaking points for a piecewise function).
Thus, we should watch for the following:
• Fractions: watch for points where the denominator is 0. Be careful that
functions which might not appear to be a fraction may be a fraction. For
sin x
.
example, tan x =
cos x
13
• Piecewise functions: When studying continuity of piecewise functions,
one should first study the continuity of each piece by using the theorems
above. Then, one must also check the continuity at each of the breaking
points.
• Functions containing absolute values: Since an absolute value can
be written as a piecewise function, they should be treated like a piecewise
function.
Theorem 36 Assume f and g are continuous at a and c is a constant. Then,
the following functions are also continuous at a:
1. f + g
2. f − g
3. cf
4. f g
5.
f
if g(a) = 0
g
Theorem 37 Any polynomial function is continuous everywhere, that is on
(−∞, ∞) .
Theorem 38 Any rational function is continuous everywhere it is defined
√
Theorem 39 If n is a positive even integer, then n x is continuous on [0, ∞)
√
Theorem 40 If n is a positive odd integer, then n x is continuous on (−∞, ∞)
Theorem 41 The trigonometric functions and their inverses, the natural logarithm functions and the exponential functions are continuous wherever they are
defined.
• One of the important properties of continuous functions is the intermediate value theorem.
Theorem 42 (Intermediate Value Theorem) Suppose f is continuous in the closed interval [a, b] and let N be a number strictly between f (a)
and f (b). Then, there exists a number c in (a, b) such that f (c) = N
• This theorem is often used to show an equation has a solution in an
interval.
Example 43 Show that the equation x2 − 3 = 0 has a solution between 1
and 2.
Let f (x) = x2 − 3. Clearly, f is continuous since it is a polynomial.
f (1) = −2, f (2) = 1. So, 0 is between f (1) and f (2). By the intermediate value theorem, there exists a number c between 1 and 2 such that
f (c) = 0.
14
• Another important property of continuous functions is that it makes computing limits easier. By the third condition in the definition of limits, we
know that if f is continuous at a then lim f (x) = f (a). This means that
x→a
if f is continuous at a,to compute lim f (x), we just ”plug in” a in the
x→a
function.
2.2
Examples
x2 + 5
continuous at x = 2?
x−1
f is a rational function, so it is continuous where it is defined. Since it is
defined at x = 2, it is continuous there.
1. Is f (x) =
x2 + 5
continuous at x = 1?
x−1
f is not defined at x = 1, hence it is not continuous at 1.
x + 1 if x ≥ 2
3. Let f (x) =
. Is f continuous at x = 2, x = 3?
x − 2 if x < 2
2. Is f (x) =
• Continuity at x = 2
Since f is a piecewise function and 2 is a breaking point, we need to
investigate the continuity there. We check the three conditions which
make a function continuous. First, f is defined at 2. Next, we see
that lim− f (x) = 0 and lim+ f (x) = 3. Thus, lim f (x) does not
x→2
x→2
x→2
exist. f is not continuous at 2.
• Continuity at 3.
When x is close to 3, f (x) = x + 1, which is a polynomial, hence it
is continuous.
4. Find where ln (x − 5) is continuous.
The natural logarithm function is continuous where it is defined. ln (x − 5)
is defined when
x−5>0
x>5
So, ln (x − 5) is continuous on (5, ∞).
x−1
is continuous.
x+2
Since f is a rational function, it is continuous where it is defined that is
for all reals except x = −2.
x + 1 if x > 2
6. Find where f (x) =
is continuous.
x − 2 if x < 2
When x > 2, f (x) = x + 1 is a polynomial, so it is continuous. When
5. Find where f (x) =
15
x < 2, f (x) = x − 2 is also a polynomial, so it is continuous. At x = 2, f
is not defined, so it is not continuous. Thus, f is continuous everywhere,
except at 2.
2.3
Sample problems:
1. Do # 3, 5, 11, 25 on pages 128, 129
⎧
if
x<0
⎨ x
if 0 < x ≤ 2 continuous at 0, 1, 2.
x2
2. Is f (x) =
⎩
8 − x if
x>2
(answer: not continuous at 0 and 2, continuous at 1)
Explain.
2x + 1
is continuous. (answer: Continuous for
x2 + x − 6
all reals except at x = 2 and x = −3)
3. Find where f (x) =
4. Find where ln (2 − x) is continuous. (answer: continuous on (−∞, 2))
5. Use the intermediate value theorem to show that x5 − 2x4 − x − 3 = 0 has
a solution in (2, 3).
6. Use the intermediate value theorem to show
√ that there is a number c such
that c2 = 3. This proves the existence of 3.
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3
Limit Laws
Suppose that c is a constant, n is a positive integer, and the limits lim f (x) and
x→a
lim g(x) exist. Then:
x→a
1. lim [f (x) + g(x)] = lim f (x) + lim g(x)
x→a
x→a
x→a
2. lim [f (x) − g(x)] = lim f (x) − lim g(x)
x→a
x→a
x→a
3. lim cf (x) = c lim f (x)
x→a
x→a
4. lim [f (x)g(x)] = lim f (x) lim g(x)
x→a
x→a
x→a
lim f (x)
f (x)
= x→a
if lim g(x) = 0
x→a g(x)
lim g(x) x→a
5. lim
x→a
n
n
6. lim [f (x)] = lim f (x)
x→a
x→a
7. lim c = c
x→a
8. lim x = a
x→a
9. lim xn = an
x→a
√
√
10. lim n x = n a If n is even, we must also have a ≥ 0
x→a
11. lim n f (x) = n lim f (x) If n is even, we must also have lim f (x) ≥ 0
x→a
x→a
x→a
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