Limits and Continuity
One of the major tasks in analysis is to classify a function by how “nice” it is. Of course,
“nice” often depends upon what you wish to do or model with the function. Below is a list of some
of the function classifications we will encounter in this course:
1.
continuity
2.
uniform continuity
3.
Lipschitz
4.
differentiable
5.
monotonic & bounded variation
6.
(Riemann-Stieltjes) integrable
We begin with an investigation of continuous functions. The mathematical meaning of continuous
is consistent with the standard English usage of the word. Continuous: Uninterrupted in time,
sequence, substance, or extent. (Mindscape Complete Reference Library CD-ROM)
Definition 25
Suppose that a function f is defined on an open interval I containing the point a. We say that the
function f is continuous at a iff for each
there exists a
such that
whenever
.
Notes:
1.
2.
Regard
as the “maximal output error” and
depends on the output error , the function
5.
, and the point a; that is,
]
3.
4.
as the corresponding “maximal input error”
]
In general, continuity is difficult to establish via Definition 25.
KEJ 34
.
6.
The above definition of continuity is essentially due to Bernard Bolzano (1781 - 1848) and
appeared in an essay he wrote in 1817 although it was largely overlooked at the time.
Augustin-Louis Cauchy (1789 - 1857) is often given credit for the first reasonable
characterization of the notion of continuity for his work in 1821.
7.
Negation of the above definition: A function
some
and every
there exists
and
.
Example 13
1.
Show that
is continuous at x = 4.
2.
Show that
is continuous at x = a.
KEJ 35
is not continuous at
such that
if for
3.
Show that
4.
Show that
5.
Show that
is continuous at x = 3.
is continuous at x = 9.
is continuous at x = a (a … 0).
Example 14
Define
Determine the points, if any, where f is continuous.
Illustration 9
(1)
Find m so that
is continuous at
(2)
.
Find b so that
is continuous at
.
Answers:
KEJ 36
Answer: x = 0, 1.
Example 15
Suppose that
where
of
is defined as follows:
is a fraction in lowest terms. Then
. The graph
over [0,1] is shown below:
The function
is continuous at each rational x and discontinuous at each rational x. We
prove this fact later.
Illustration 10
(1)
Prove that
is continuous at
KEJ 37
.
Solution
Let
be given. Choose
since the function
fact!). We conclude that
(2)
. Then for
for
we have that
(do the calculus to verify this critical
is continuous at
Prove that the rational function
is continuous at
. €
.
Solution
Let
be given. Choose
. Then for
KEJ 38
we have that
since
for
conclude that
(again, do the calculus to verify this fact!). We
. €
is continuous for
Illustration 11
(1)
Explain why the function
(2)
Explain why the function
fails to be continuous at
is everywhere continuous including at
(3)
.
.
Explain why the function
fails to be continuous at
no matter what value we assign to
KEJ 39
.
Illustration 12
Prove that
is continuous at
.
Solution (sketch)
Let
. Let x be such that
be given. Choose
that for all such
. We observe
(graph this difference quotient on the
that
interval (1.8,2.2) to verify the stated bound!). Consider
We conclude that
is continuous at
.
The next result gives a characterization of continuity in terms of sequences; it states that
continuous functions carry convergent sequences to convergent sequences.
Theorem 26 - Sequential Criterion for Continuity
Let f be defined on an open interval I = (b,c) containing the point a (that is,
following are equivalent:
1.
f is continuous at a;
KEJ 40
) . Then the
2.
f I with
if
, then
.
Proof
(Y) Suppose f is continuous at x = a. Let
f I with
be given and let
Because f is continuous at x = a, there exists a
.
so that
whenever
.
Because
there exists N 0 ù so that if
we have
. Thus,
,
. It follows that for
.
(Z) In this case, we assume that condition 1 is false and we show that condition 2 is also false.
Because f is discontinuous at x = a, there exists at least one
there exists x with
construct a sequence
converges to a while
and
so that for all possible
. By choosing
f I with
and
does not converge to
(n = 1, 2, 3, ... ) we
. Thus,
. >
The previous result is probably most useful in showing that a particular function is not continuous
at a particular point x = a.
Example 16
We define the Dirichlet function as follows
Show that the Dirichlet function is nowhere continuous.
KEJ 41
Solution
Let a 0 ú.
(Case 1) Suppose that a 0 Q. Define a sequence
(Note that
as follows
is rational when n is odd and is irrational when n is even.) Then
oscillates between 0 and 1 and, hence, it fails to converge to
but
.
(Case 2) Exercise. >
Example 17
Suppose that the function f is continuous at each point in the interval
every rational number r. Prove that
for all x in
with
for
.
Solution
Let x be any number in
. Find a sequence of rational numbers
. Since f is continuous at x, Theorem 26 implies that
in
. We conclude
We now extend the pointwise notion of continuity to a setwise formulation.
Definition 27
Suppose
1.
a 0 D;
2.
for each
. The function f is continuous at a with respect to D iff
there exists
such that
KEJ 42
such that
.
whenever
and
.
A function is said to be continuous on D if it is continuous with respect to D at every point of D.
A special case of the above is when D is a closed bounded interval [a,b]. In this case the definition
automatically takes of care of the “one-sided” continuity at the endpoints. (Explain why?)
Example 18
1.
Polynomial functions are continuous on ú.
2.
Rational functions are continuous on their domain of definition.
3.
Power functions such as
are continuous on their domain of definition.
Example 19
Consider the function
point
which consists of the straight line segments connecting the
with the two points
and
and
. The graph of
where
is shown below:
The above is clearly continuous away from zero given that it is piecewise linear. To see that it is
continuous at zero we note that
and so continuity there follows by choosing
.
KEJ 43
Point: The function
is continuous on [-1,1] yet there is no way to sketch the graph of
without lifting the pencil from the page. (The problem is to connect (0,0) to (1,1) or (-1,-1)
to (0,0).)
Definition 28
Suppose that f is defined on
for some
approaches a is L, written
, iff for each
. Then the limit of f(x) as x
there exists a
so that
whenever
.
According to Definition 28 the function f(x) under consideration need not be defined at x = a.
Further, if it is defined at x = a, its behavior at x = a is of no consequence with respect to the limit
process.
Example 20
Let a, c 0 ú.
1.
Show that
.
(Any ä works!)
2.
Show that
.
(Choose ä = å.)
3.
Show that
.
(Choose
Example 21
Show that
if
.
KEJ 44
.)
Solution
Let
be given. Choose
(i) Suppose x < 4 with
. Then
(ii) Suppose that x > 4 with
So, if
. We consider two cases: (i) x > 4 and (ii) x < 4.
, then
. Consider
. Then
. Consider
. €
and so
Note that in the above the behavior of the function
at x = 4 was of no interest to us and its
value at x = 4 played no role in the solution of the problem..
KEJ 45
Below we state as a theorem the standard Calculus I characterization of continuity in terms
of limits.
Theorem 29
Suppose that f is defined on an open interval I containing the point x. Then f is continuous at
x = a if and only if
.
One useful consequence of Theorem 29 is that it is typically the case that for each theorem we prove
concerning limits there is a corresponding result for continuity.
Example 22 (See Example 15)
Define f : ú 6 ú by
for all a 0 ú.
a.
Explain why
b.
Use part a and Theorem 29 to conclude that f is continuous precisely on the irrational
numbers.
Theorem 30
Suppose that f is defined on
for some r > 0. If
, then L = M.
KEJ 46
and
Example 23
Suppose that g(x) is bounded on
that
(some r > 0). If
, then show
.
Solution
for x in
Let å > 0 be given. Suppose that
. Since
,
there exists ä > 0 so that
whenever
.
Let x be a number such that
. >
We conclude that
It is not required in the above that
on
. Then
exist; the function g need only be defined and bounded
.
The next example is quite useful!
KEJ 47
Example 24
1.
and f is continuous at L, show that
If
.
Solution
Let å > 0 be given. Since f is continuous at L,
(Theorem 29). So, there exists
such that
whenever
.
Because
, there exists ä > 0 so that
whenever
.
Therefore, for x with
we have
.
. >
Hence,
In view of the above, we say that a continuous function commutes with limits.
2.
Evaluate
.
Solution
From Calculus I we recall that
. Set
in particular, f is continuous at 1. By part 1, we have
KEJ 48
. Then f is continuous on
and,
Theorem 31 - Algebra of Limits
Suppose that
and
.
1.
;
2.
3.
;
provided M … 0;
provided M … 0.
4.
Proof
1. Let å > 0 be given. Since
, there exists a
whenever
.
Similarly, since
, there exists a
KEJ 49
so that
so that
whenever
.
. For x such that
Choose
we have
Therefore,
. (The subtraction case is similar and is
omitted.)
2. Let å > 0 be given. Since
, there exists a
so that
whenever
.
Thus, f is bounded on
that
Hence,
. Because
and
and
. Therefore, by Example 17,
.
KEJ 50
, it follows
3. To obtain the desired results set
and apply Example 18.
4. Follows from parts 2 and 3. >
By a standard induction argument, the results of Theorem 31 extend to finite sums and products.
Example 25
If
and
both exist, does it follow that
exists?
Solution
No, consider
and
.
The next result is an immediate consequence of the Algebra of Limits Theorem and the
definition of a continuous function.
Corollary 32
Suppose the functions f and g are continuous at x = a.
1.
2.
is continuous at x = a.
is continuous at x = a.
3.
is continuous at x = a provided
4.
is continuous at x = a provided that
.
.
We note that Example 16 and Corollary 32 now establishes Example 15. Combining Example 20
with Corollary 32 we have the following result on composite functions.
KEJ 51
Corollary 33
If g is continuous at a and f is continuous at g(a), then
is continuous at x = a.
We now state and prove a result that goes by a variety of names including the Sandwiching
Theorem, Pinching Theorem, and Squeeze Theorem.
Theorem 34 - Sandwiching Theorem
Suppose that f, g, and h are all defined on the set
for all x 0 S and
1.
2.
then
(some r > 0). If
,
.
Proof - Optional
Let å > 0 be given. Since
there
and
so that
whenever
and
whenever
.
Choose
. Then for real numbers x with
.
Thus,
whenever
KEJ 52
we have
.
. >
Hence,
Example 26
for x 0 ú. Show that
Suppose that
.
Solution
Since
and
,
, the Sandwiching Theorem implies that
. Because
.
We now consider one-sided limits.
Definition 35
Suppose that f is defined on (a-r,a) (some r > 0). Then the limit of f(x) as x approaches a from the
left is L, written
. iff for each å > 0 there exists a ä > 0 so that
whenever
.
There is an analogous definition for right-hand limits. The Algebra of Limits Theorem holds for
both left-hand and right-hand limits. Further, there are versions of the Sandwiching Theorem for
both left-hand and right-hand limits. The following notations are convenient:
KEJ 53
provided the limits exist.
The next result characterizes two-sided limits in terms of one-sided limits. The result is most
often used to show that a particular function does not have a two-sided limit at a specific point.
Theorem 36
Suppose that f is defined on
(some r > 0). Then
iff
.
Proof (Sketch)
(Y) Given å > 0, the same value of ä works for all limits.
. >
(Z) Let å > 0 be given. Choose
In view of Theorems 29 and 36, a function f is continuous at x = a iff
.
Definition 37
1.
Suppose that f is defined on (0,4) and we set
. Then
iff
. Then
iff
.
2.
Suppose that f is defined on (-4,0) and we set
.
We say that the horizontal line y = L is an asymptote for the graph of
or
.
KEJ 54
iff either
Example 27
(k 0 ú) since
1.
2.
Show that
.
Solution
Set
. Then
. Because
for all x,
, by the Sandwich Theorem for right-hand limits, it follows that
. The desired results concerning f is established.
for x > 0, the Sandwiching Theorem for
In Example 25, we note that since
limits at infinity also provides the desired result.
Theorem 38
Suppose that f is continuous on the interval
then there exists
such that
. If
.
Proof
WLOG, assume
. Define
.
KEJ 55
and
have opposite signs,
Since
and S is bounded above by b. By the Least Upper Bound Property
,
of ú, S has a least upper bound, say
such that
, it follows that
, we see that
By the continuity of f at c,
and so
there exists
. Because f is continuous at c, we have that
. Clearly,
. Because each
Given that
. By Theorem 11, for each
.
. Construct a sequence
. Because
where
and
.
, we conclude that
.€
. It must be the case that
Example 28
1.
Show that
2. `
Show that
has a root / zero / x-intercept on
has a root / zero / x-intercept on
.
.
Theorem 39 - Intermediate Value Theorem
Suppose f is continuous on the interval
, then there exists a
. If k is any number between
such that
.
Proof
Apply Theorem 38 to the function
.
The next two results follow from the Bolzano-Weierstrass Theorem.
Theorem 40
Suppose f is continuous on the interval
. Then f is bounded there.
KEJ 56
and
Theorem 41 - Extreme-Value Theorem
Suppose f is continuous on the interval
. Then there are points
such that
and
.
KEJ 57