Limits of Functions

In this chapter, we study another notion of convergence that is surely familiar to the reader, namely, the limit of a function at a given point. After introducing the precise definition of the limit of a function, and working through some examples, we will relate limits of functions with limits of sequences resulting in the Sequential Criterion for Limits (Theorem 4.1.11). In this chapter, when not explicitly stated, the letter

A

will denote a subset of

R

Before we can give the definition of the limit of a function, we need the notion of a cluster point of a set.

A number

c \in R

is called a cluster point of

A

if for any given

δ > 0

there exists at least one point

x \in A

, with

x \neq c

, such that

| x - c | < δ

Hence,

c

is a cluster point of

A

if there are points in

A

that are arbitrarily close to

c

. In general, a cluster point of

A

is not necessarily an element of

A

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

A point

c

is a cluster point of

A

if and only if there exists a sequence

(x_{n})

A

such that

x_{n} \neq c

and

lim_{n \to \infty} x_{n} = c

Let

c

be a cluster point of

A

and let

δ_{n} = \frac{1}{n}

for

n \in N

. Then by definition of a cluster point, there exists

x_{n} \in A

x_{n} \neq c

, such that

| x_{n} - c | < δ_{n}

. Since

δ_{n} \to 0

then

x_{n} \to c

. To prove the converse, suppose that

x_{n} \to c

x_{n} \neq c

and

x_{n} \in A

for

n \in N

. Then by convergence of

(x_{n})

c

, for any

δ > 0

there exists

K \in N

such that

| x_{K} - c | < δ

. Since

x = x_{K} \in A

, this proves that

c

is a cluster point of

A

Below are some examples of cluster points for a given set:

Consider the set $A = [0, 1]$ . Every point $c \in A$ is a cluster point of $A$ .
On the other hand, for $A = (0, 1]$ , the point $c = 0$ is a cluster point of $A$ but does not belong to $A$ .
For $A = {\frac{1}{n} | n \in N}$ , the only cluster point of $A$ is $c = 0$ .
A finite set does not have any cluster points.
The set $A = N$ has no cluster points.
Consider the set $A = Q \cap [0, 1]$ . By the Density theorem, every point $c \in [0, 1]$ is a cluster point of $A$ .

We now give the definition of the limit of a function

f : A \to R

at a cluster point

c

A

Consider a function

f : A \to R

and let

c

be a cluster point of

A

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

L \in R

such that for any given

ε > 0

there exists

δ > 0

such that if

x \in A

and

0 < | x - c | < δ

then

| f (x) - L | < ε

. In this case, we write that

lim_{x \to c} f (x) = L

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

f

does not converge at

c

then we say that

f

diverges at

c

Another short-hand notation to denote that

converges to

. By definition, if

, then for any

there exists a

such that for all

not equal to

it holds that

A function

can have at most one limit at

Suppose that

and

, and let

. Then there exists

such that

and

, for all

satisfying

. Then if

then

Since

is arbitrary, Theorem 2.2.7 implies that

Consider the function

with domain

. Prove that

We begin by analyzing the quantity

Hence, if

then

Thus, given

we let

and thus if

then

. Thus, by definition,

Consider the function

with domain

. Prove that

We have that

Let

be arbitrary and let

. Then if

then

. Hence, if

then

Thus, by definition,

Consider the function

with domain

. Prove that for any

We first note that

By the triangle inequality,

and therefore

We now need to analyze how large

can become when

is say within

. To be concrete, suppose that

. Hence, if

then

Therefore, if

it holds that

Now suppose that

is arbitrary and let

. Then if

then

and therefore

This proves, by definition, that

for any

Consider the function

with domain

. Prove that

We first note that

is indeed a cluster point of

. Now,

We now obtain a bound for

when

is close to

. Suppose then that

. Then

and therefore,

, which implies that

. Similarly, if

then

and therefore

, which implies that

. Therefore, if

then

Suppose now that

is arbitrary and let

. If

then from our analysis above it follows that

. Therefore, if

then

This proves that

Consider the function

defined as

Prove that

then

and if

then

Therefore, for all

it holds that

since

. Thus, given

let

and thus if

then

This proves 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

(with

for all

) the sequence

converges to

Suppose that

. Let

be a sequence in

converging to

, with

for all

. We must prove that the sequence

converges to

. To that end, let

be arbitrary. Then, by convergence of

, 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 must 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, negating the definition of the limit of a function, there exists

such for all

there exists

such that

and

. Then, let

for

. Then there exists

such that

and

. Since

then

but clearly

does not converge to

. This ends the proof.

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

for all

, and such that

does not converge to

Note that in Corollary 4.1.12, if the sequence

diverges then by definition it 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

for all

. If

and

converge but

then

does not have a limit at

Prove that

does not exist.

Consider

, which clearly converges to

and

for all

. Then

which is unbounded and thus does not converge. Thus, by Corollary 4.1.12,

does not exist.

Prove that

does not exist.

Let

with domain

. 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 sign function, denoted by

, is defined as

Prove that

does not exist.

Consider the sequence

. Then

and

for all

. Now,

and thus

does not converge. Therefore, by Corollary 4.1.12, the function

has no limit at

Exercises

Use the definition of the limit of a function to prove that the following limits do indeed hold.

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.

Recall that the function

is defined as follows:

Limit Theorems

In this section, we establish basic limit theorems for limits of functions. The reader should compare the results of this section with Section 3.2 where we established limit theorems for sequences. In fact, thanks to the sequential criterion for limits of functions (Theorem 4.1.11), all of the theorems in this section can be proved using limits of sequences. To begin, we first show that if

has a limit at

then

satisfies a local boundedness property at

. Let us first define then what it means for a function to be locally bounded at a given point.

Consider a function

and let

be a cluster point of

. We say that

is bounded locally at if there exists

and

such that if

then

Consider 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 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 state and prove some limit laws for functions. Let

be functions and define the functions

, and

as follows:

where for

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

. Instead, notice that if

then

. Hence, the functions

and

are equal at every point in

. It is clear that

and therefore it follows that also

Let

be a function and let

be a cluster point of

. Suppose that

has limit

. If

for all

then

We prove the contrapositive. Suppose then that

. Let

be such that

. Then since

, there exists

such that if

then

. Hence,

for some

. We give another proof using the sequential criterion for limits. To that end, if

converges to

then for any sequence

converging to

, we have that

. Now

and therefore

from our results on limits of sequences (Theorem 3.2.7).

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

from our results on limits of sequences (Theorem 3.2.7).

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

with

for all

. Then, by the sequential criterion,

By assumption, it holds that

for all

, 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

, and thus by transitivity it follows 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 of the limit of a function, 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