Limits | An Introduction to Real Analysis

You are familiar with computing limits of functions from calculus. As you may recall, a function

has limit

at a point

if the outputs

are arbitrarily close to

provided the inputs

are sufficiently close to

. Although this notion is intuitive, we will give a precise definition of the limit of a function and we will relate the definition with limits of sequences.

Limits of Functions

Recall that when computing

the function

need not be defined at

. The reason for this is that many limits of interest arise when

is not defined at

, for example when computing derivatives. However, in order for the limit notion to make sense, it is necessary that

be defined for points arbitrarily close to

. This motivates the following definition.

Let

. The point

is called a cluster point of

if for any given

there exists at least one point

not equal to

such that

Notice that a cluster point of a set

may or may not be a point in

. The idea of a cluster point

is that there are points in

that are arbitrarily close to

. Naturally, cluster points can be characterized using limits of sequences.

A point

is a cluster point of

if and only if there exists a sequence

such that

and

Let

be a cluster point of

and let

. Then, there exists

, such that

. Since

we have that

. Now suppose that

and

, and

is in

. Then for any

there exists

such that

for all

. This proves that

is a cluster point of

Here are some examples of cluster points:

Consider the set . Every point is a cluster point of .
On the other hand, for , the point is a cluster point of but does not belong to .
For , the only cluster point of is .
A finite set does not have any cluster points.
The set has no cluster points.
Consider the set . By the Density theorem, every point is a cluster point of .

We now give the definition of the limit of a function

at a cluster point

. In this chapter, the letter

will denote a subset of

Consider a function

and let

be a cluster point of

. We say that has limit at , or converges at , if there exists a number

such that for any given

there exists

such that if

and

then

. In this case, we write that

and we say that converges to at c, or that has limit at . If

does not converge at

then we say that it diverges at

Another short-hand notation to say that

converges to

. By definition, if

, then for any

neighborhood of

, there exists a

neighborhood of

such that for all

not equal to

it is true that

A function

can have at most one limit at

Suppose that

and

. Let

. Then there exists

such that

and

, for

. Then if

then

Since

is arbitrary, Theorem 2.2.7 implies that

Prove that

We begin by analyzing the quantity

Hence, if

then

Prove that

We have that

Let

be arbitrary and let

. Then if

then

. Hence, if

then

Prove that

We have that

By the triangle inequality,

and therefore

We now obtain a bound for

when

is say within

. Hence, suppose that

. Then

and therefore

. Suppose that

is arbitrary and let

. Then if

then

and therefore

Prove that

We have that

We now obtain a bound for

when

is close to

. Suppose then that

. Then

. Therefore,

, and therefore

. Similarly, if

then

and therefore

. Therefore, if

then

Suppose that

is arbitrary and let

. If

then from above we have that

, and clearly

. Therefore, if

then

This ends the proof.

Prove that

The following important result states that limits of functions can be studied using limits of sequences.

Let

be a function and let

be a cluster point of

. Then

if and only if for every sequence

converging to

the sequence

converges to

(with

Suppose that

. Let

be a sequence in

converging to

, with

. Let

be given. Then there exists

such that if

then

. Now, since

, there exists

such that

for all

. Therefore, for

we have that

. This proves that

. To prove the converse, we prove the contrapositive. Hence, we show that if

does not converge to

then there exists a sequence

(with

) converging to

but the sequence

does not converge to

. Assume then that

does not converge to

. Then there exists

such for all

there exists

such that

and

. Let

. Then there exists

such that

and

. Hence

but

does not converge to

The following theorem follows immediately from Theorem 4.1.11.

Let

be a function and let

be a cluster point of

and let

. Then

does not converge to

if and only if there exists a sequence

converging

, with

, and such that

does not converge to

Note that in the previous corollary, if the sequence

diverges then it clearly does not converge to any

and then

does not have a limit at

. When applicable, the following corollary is a useful tool to prove that a limit of a function does not exist.

Let

be a function and let

be a cluster point of

. Suppose that

and

are sequences in

converging to

, with

and

. If

and

converge but

then

does not have a limit at

Prove that

does not exist.

Consider

, which clearly converges to

and

. Then

which does not converge.

Prove that

does not exist.

Let

. Consider the sequence

. It is clear that

and

for all

. Now

and therefore

does not converge. Therefore,

has no limit at

. In fact, for each

, consider the sequence

. Clearly

and

for all

. Now,

. Hence,

converges to

. This shows that

oscillates within the interval

approaches

The

function is defined as

Prove that

does not exist.

Let

. Then

and

for all

. Now,

. Clearly,

does not converge and thus

has no limit at

Exercises

Use the

definition of the limit of a function to verify the following limits.

Let

, let

, and suppose that

is a cluster point of

. Suppose that there exists a constant

such that

for all

. Prove that

Consider the function

Prove that

Let

be defined as follows:

Prove that has a limit at .
Now suppose that . Prove that has no limit at .
Define by . Prove that has a limit at any .

Hint: The Density Theorem will be helpful for (b). In particular, the Density Theorem implies that for any point

, there exists a sequence

of rational numbers such that

, and that there exists a sequence

of irrational numbers such that

Use any applicable theorem to explain why the following limits do not exist.

The sgn function is defined as follows:

Limit Theorems

We first show that if

has a limit at

then it is locally bounded at

. We first define what it means to be locally bounded.

Let

and let

be a cluster point of

. We say that

is bounded locally at if there exists

and

such that if

then

Let

be a function and let

be a cluster point of

. If

exists then

is bounded locally at

Let

and let

be arbitrary. Then there exists

such that

for all

such that

. Therefore, for all

and

we have that

then let

and if

then let

. Then

for all

such that

, that is,

is bounded locally at

Consider again the function

defined on the set

. Clearly,

is a cluster point of

. For any

and any

let

be such that

. Then

, that is,

. Since

was arbitrary, this proves that

is unbounded at

and consequently

does not have a limit at

We now consider Limit Laws for functions. Let

be functions and define

where for the ratio

we require that

for all

Let

be functions and let

be a cluster point of

. Suppose that

and

. Then

, if

The proofs are left as an exercises. (To prove the results, use the sequential criterion for limits and the limits laws for sequences).

Let

be functions and let

be a cluster point of

. If

exists for each

then

is a polynomial function then

for every

. If

is another polynomial function and

in a neighborhood of

and

then

Prove that

We cannot use the Limit Laws directly since

. Now, if

then

. Hence, the functions

and

agree at every point in

. It is clear that

. Let

and let

be such that if

then

. Then for

we have that

. Hence

Let

be a function and let

be a cluster point of

. Suppose that

has limit

. If

for all

then

We prove the contrapositive. Suppose that

. Let

be such that

. There exists

such that if

then

. Hence,

for some

. An alternative proof: If

converges to

then for any sequence

converging to

, we have that

. Now

and therefore

from our results on limits of sequences.

Let

be a function and let

be a cluster point of

. Suppose that

for all

and suppose that

. Then

We have that

and therefore by Theorem 4.2.8 we have that

. Similarly, from

we deduce that

. From this we conclude that

. An alternative proof: Since

, for any sequence

with

, we have that

. Clearly,

and therefore

The following is the Squeeze Theorem for functions.

Let

be functions and let

be a cluster point of

. Suppose that

and

. If

for all

, then

Let

be a sequence in

converging to

. Then

. Clearly

, and therefore by the Squeeze Theorem for sequences we have that

. This holds for every such sequence and therefore

Let

Show that

We end this section with the following theorem.

Let

be a function and let

be a cluster point of

. Suppose that

. If

then there exists

such that

for all

Choose

so that

, take for example

. Then there exists

such that

for all

Exercises

Let

and suppose that

is a cluster point of

. Suppose that at

converges to

and

converges to

. Prove that

converges to

in two ways: (1) using the

definition, and (2) using the sequential criterion for limits.

Give an example of a set

, a cluster point

, and two functions

such that

exists but

does not exist.

Give an example of a function

that is bounded locally at

but does not have a limit at

. Your answer should not be in the form of a graph.

Let

be a function that is bounded locally at

and suppose that

converges to

. Prove that