Some Basic Properties Of Limits

In this section, we cover some limit theorems that will be needed in later proofs. If you are less interested in the formal theoretical foundations and proofs, this section may be safely skipped on a first reading.

Quick Reference

Property	Statement	Significance
Limit Uniqueness	If $lim_{x \to a} f (x) = L_{1}$ and $L_{2}$ , then $L_{1} = L_{2}$	A function cannot approach two different limits at the same point.
One-Sided Equivalence	$lim_{x \to a} f (x) = L ⟺ lim_{x \to a^{-}} f (x) = lim_{x \to a^{+}} f (x) = L$	Two-sided limits exist if and only if left and right limits exist and agree.
Infinitesimal Representation	$lim_{x \to a} f (x) = L ⟺ f (x) = L + α (x)$ with $lim_{x \to a} α (x) = 0$	Expresses limit behavior as a constant plus an infinitesimal.
Local Boundedness	$lim_{x \to a} f (x) = L ⟹ \| f (x) \| < K$ near a	A function with a real limit is locally bounded near the limit point.
Sign Preservation	$lim_{x \to a} f (x) = L \neq 0 ⟹ f (x)$ shares the sign of $L$ near a	Non-zero limits preserve sign in a deleted neighborhood of a.

In the first section of this chapter, using intuition, we explained that the limit of a function is always unique. After providing the rigorous definition of a limit, we can prove it.

Theorem: If $f (x)$ approaches a limit as $x \to a$ , then this limit is unique.

Proof

Suppose $lim_{x \to a} f (x) = L_{1}$ and $lim_{x \to a} f (x) = L_{2}$ where $L_{1} \neq L_{2}$ .

Let $ϵ = \frac{1}{2} | L_{1} - L_{2} |$ . Because $lim_{x \to a} f (x) = L_{1}$ , according to the definition, there exists some $δ_{1} > 0$ such that $| f (x) - L_{1} | < ϵ$ whenever $0 < | x - a | < δ_{1}$ . Similarly because $lim_{x \to a} f (x) = L_{2}$ , there exists some $δ_{2} > 0$ such that $| f (x) - L_{2} | < ϵ$ whenever $0 < | x - a | < δ_{2}$ .

Now using the triangle inequality $| A + B | \leq | A | + | B |$ , we can write:

\begin{aligned} |L_1 - L_2| = |(L_1 - f(x)) + (f(x) - L_2)| &\le |L_1 - f(x)| + |f(x) - L_2| \\ &< \epsilon + \epsilon = |L_1 - L_2| \end{aligned}

whenever $0 < | x - a | < δ = min {δ_{1}, δ_{2}}$ . However, $| L_{1} - L_{2} | < | L_{1} - L_{2} |$ is impossible because we assumed $L_{1} \neq L_{2}$ so $L_{1} - L_{2} \neq 0$ . This contradiction proves the theorem.

Theorem: $lim_{x \to a} f (x)$ exists and is equal to $L$ if and only if $lim_{x \to a^{-}} f (x)$ and $lim_{x \to a^{+}} f (x)$ both exist and are equal to $L$ . That is,

lim_{x \to a} f (x) = L ⟺ lim_{x \to a^{-}} f (x) = lim_{x \to a^{+}} f (x) = L .

Proof

We prove the bi-implication by showing both directions separately.

Direct Implication ( $⟹$ ): Assume that $lim_{x \to a} f (x) = L$ . By definition, for every $ϵ > 0$ , there exists a $δ > 0$ such that $0 < | x - a | < δ ⟹ | f (x) - L | < ϵ$ .

If $a - δ < x < a$ , then $0 < a - x = | x - a | < δ$ . Hence, $| f (x) - L | < ϵ$ , which shows that $lim_{x \to a^{-}} f (x) = L$ .
If $a < x < a + δ$ , then $0 < x - a = | x - a | < δ$ . Hence, $| f (x) - L | < ϵ$ , which shows that $lim_{x \to a^{+}} f (x) = L$ .

Reverse Implication ( $⟸$ ): Assume that

lim_{x \to a^{-}} f (x) = L

and

lim_{x \to a^{+}} f (x) = L

. Let

ϵ > 0

be given.

Since $lim_{x \to a^{-}} f (x) = L$ , there exists a $δ_{1} > 0$ such that $a - δ_{1} < x < a ⟹ | f (x) - L | < ϵ$ .
Since $lim_{x \to a^{+}} f (x) = L$ , there exists a $δ_{2} > 0$ such that $a < x < a + δ_{2} ⟹ | f (x) - L | < ϵ$ .

Choose

δ = min {δ_{1}, δ_{2}} > 0

. Now suppose

0 < | x - a | < δ

. This condition implies either

a - δ < x < a

a < x < a + δ

If $a - δ < x < a$ , then $a - δ_{1} \leq a - δ < x < a$ , so $| f (x) - L | < ϵ$ .
If $a < x < a + δ$ , then $a < x < a + δ \leq a + δ_{2}$ , so $| f (x) - L | < ϵ$ .

In either case,

| f (x) - L | < ϵ

whenever

0 < | x - a | < δ

. Thus,

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

A function is called infinitesimal as $x \to a$ if $lim_{x \to a} α (x) = 0$ . The following theorem states that the limit of a function $f (x)$ as $x \to a$ is a number $L$ if $f (x)$ can be written as $L$ plus an infinitesimal as $x \to a$ . Conversely, if $f (x)$ is written as the sum of a number $L$ and an infinitesimal as $x \to a$ , then $lim_{x \to a} f (x) = L$ .

Theorem: The function f approaches $L$ as x approaches a if and only if $f (x) = L + α (x)$ , where $lim_{x \to a} α (x) = 0$ .

Proof

$(⟹)$ First, assume that $lim_{x \to a} f (x) = L$ . Define the function $α (x)$ by $α (x) = f (x) - L$ . Rearranging this equation gives $f (x) = L + α (x)$ . Taking the limit as $x \to a$ on both sides yields:

\begin{aligned} \lim_{x\to a} \alpha(x) &= \lim_{x\to a} [f(x) - L] \\ &= \lim_{x\to a} f(x) - \lim_{x\to a} L \\ &= L - L = 0 \end{aligned}

Thus, $f (x)$ can be represented in the required form with an infinitesimal $α (x)$ .

$(⟸)$ Conversely, assume that $f (x) = L + α (x)$ where $lim_{x \to a} α (x) = 0$ . Taking the limit as $x \to a$ on both sides, we apply the limit laws:

\begin{aligned} \lim_{x\to a} f(x) &= \lim_{x\to a} [L + \alpha(x)] \\ &= \lim_{x\to a} L + \lim_{x\to a} \alpha(x) \\ &= L + 0 = L \end{aligned}

Therefore, $f (x)$ approaches $L$ as x approaches a.

The above theorem remains valid if a is replaced by $+ \infty$ or $- \infty$ .

The following theorem says that if the limit of a function exists as $x \to a$ , then the function does not blow up near $x = a$ , and $y$ values remain within some strip $- K < y < K$ .

Theorem: If $lim_{x \to a} f (x) = L$ , then $f (x)$ is bounded near a. That is, there are positive numbers $K$ and $δ$ such that $| f (x) | < K$ , whenever $0 < | x - a | < δ$ .

Proof

Since $lim_{x \to a} f (x) = L$ , taking $ϵ = 1$ , there is a $δ$ such that $| f (x) - L | < ϵ$ whenever $0 < | x - a | < δ$ . Then

| f (x) | = | (f (x) - L) + L | \leq | f (x) - L | + | L | < 1 + | L | .

If f is defined at a, then let $K = max {1 + | L |, | f (a) |}$ .

If $f (x) \to L$ as $x \to a$ , $f (x)$ remains close to $L$ when x is close to a. Therefore, we expect that if $L \neq 0$ , $f (x)$ has the same sign as $L$ near a, except possibly at a (see the following figure). We have to exclude the sign of $f (a)$ because what happens at $x = a$ does not influence the limit of f as $x \to a$ .

Illustration of sign preservation near a non-zero limit point L — Sign preservation near $x = a$ .

To be more specific, we have:

Theorem: If $lim_{x \to a} f (x) = L$ and $L \neq 0$ , then there exists an open interval $I$ containing a such that $f (x)$ has the same sign as $L$ for all $x \in I ∖ {a}$ .

Proof

Let $ϵ = \frac{| L |}{2} > 0$ . By the definition of a limit, there exists a $δ > 0$ such that $| f (x) - L | < \frac{| L |}{2}$ whenever $0 < | x - a | < δ$ .

The inequality $| f (x) - L | < \frac{| L |}{2}$ is equivalent to $- \frac{| L |}{2} < f (x) - L < \frac{| L |}{2}$ . Adding $L$ to all parts yields:

L - \frac{| L |}{2} < f (x) < L + \frac{| L |}{2} .

We now consider two cases:

If $L > 0$ , then $| L | = L$ , so $0 < \frac{L}{2} < f (x) < \frac{3 L}{2}$ . Thus, $f (x) > 0$ .
If $L < 0$ , then $| L | = - L$ , so $\frac{3 L}{2} < f (x) < \frac{L}{2} < 0$ . Thus, $f (x) < 0$ .

In either case, choosing

I = (a - δ, a + δ)

gives an open interval containing a where

f (x)

has the same sign as

L

for all

x \in I ∖ {a}

Exercises

Exercise 1.

Let $f (x)$ be defined by:

f(x) = \begin{cases} 2x + 1 & \text{if } x < 1 \\ kx^2 + 2 & \text{if } x > 1 \end{cases}

Find the value of $k$ such that $lim_{x \to 1} f (x)$ exists. For this value of $k$ , what is the unique value of the limit?

Answer

$k = 1$ , and the unique limit is $L = 3$ .

Solution

By the One-Sided Equivalence Theorem, $lim_{x \to 1} f (x)$ exists if and only if $lim_{x \to 1^{-}} f (x)$ and $lim_{x \to 1^{+}} f (x)$ both exist and are equal.

Left-hand limit: $lim_{x \to 1^{-}} f (x) = lim_{x \to 1^{-}} (2 x + 1) = 2 (1) + 1 = 3$ .
Right-hand limit: $lim_{x \to 1^{+}} f (x) = lim_{x \to 1^{+}} (k x^{2} + 2) = k (1)^{2} + 2 = k + 2$ .

Equating the left-hand and right-hand limits: $k + 2 = 3 ⟹ k = 1.$ When $k = 1$ , the two-sided limit exists and by the Limit Uniqueness Theorem, its unique value is $L = 3$ .

Exercise 2.

Let $f (x) = x^{2} + 3 x$ .

Show that $lim_{x \to 2} f (x) = 10$ .
Express $f (x)$ in the infinitesimal form $f (x) = 10 + α (x)$ near $x = 2$ , explicitly find $α (x)$ , and verify that $lim_{x \to 2} α (x) = 0$ .

Answer

(a) Direct substitution gives $2^{2} + 3 (2) = 10$ . (b) $α (x) = x^{2} + 3 x - 10$ , and $lim_{x \to 2} α (x) = 0$ .

Solution

(a) By direct substitution for polynomials, $lim_{x \to 2} (x^{2} + 3 x) = 2^{2} + 3 (2) = 4 + 6 = 10$ .

(b) By the Infinitesimal Representation Theorem, $f (x) = L + α (x)$ where $L = 10$ . Rearranging gives:

α (x) = f (x) - 10 = x^{2} + 3 x - 10 = (x - 2) (x + 5) .

Evaluating the limit of $α (x)$ as $x \to 2$ :

lim_{x \to 2} α (x) = lim_{x \to 2} (x^{2} + 3 x - 10) = 2^{2} + 3 (2) - 10 = 0.

Hence, $f (x) = 10 + (x^{2} + 3 x - 10)$ represents $f (x)$ as a constant plus an infinitesimal as $x \to 2$ .

Exercise 3.

Suppose $lim_{x \to 3} f (x) = 5$ . Given that $| f (x) - 5 | < 1$ whenever $0 < | x - 3 | < 0.2$ , find a positive constant K such that $| f (x) | < K$ for all x satisfying $0 < | x - 3 | < 0.2$ .
Explain why the function $g (x) = \frac{1}{x - 3}$ is not locally bounded in any deleted neighborhood of $x = 3$ .

Answer

(a) $K = 6$ (b) As $x \to 3^{+}$ , $g (x) \to + \infty$ , so $g (x)$ grows larger than any fixed constant $M$ .

Solution

(a) By the Triangle Inequality $| A | \leq | A - B | + | B |$ :

| f (x) | = | (f (x) - 5) + 5 | \leq | f (x) - 5 | + 5 < 1 + 5 = 6.

Thus, taking $K = 6$ and $δ = 0.2$ , we have $| f (x) | < 6$ whenever $0 < | x - 3 | < 0.2$ .

(b) As $x \to 3^{+}$ , $g (x) = \frac{1}{x - 3} \to + \infty$ . For any fixed radius $δ > 0$ and any positive bound $M > 0$ , we can choose $x = 3 + \frac{1}{M + 1} \in (3, 3 + δ)$ such that $g (x) = M + 1 > M$ . Therefore, no constant $K$ can bound $| g (x) |$ near $x = 3$ , showing that $g (x)$ is not locally bounded near $x = 3$ .

Exercise 4.

Suppose $lim_{x \to 2} f (x) = - 4$ . Prove using the definition of a limit (or the Sign Preservation Theorem) that there exists a deleted neighborhood of $x = 2$ where $f (x) < - 2$ (and thus $f (x)$ is strictly negative).

Answer

$f (x) < - 2 < 0$ for all $0 < | x - 2 | < δ$ .

Solution

Here $L = - 4 < 0$ . Choose $ϵ = \frac{| L |}{2} = \frac{4}{2} = 2$ .
By the definition of a limit, there exists a $δ > 0$ such that whenever $0 < | x - 2 | < δ$ :

| f (x) - (- 4) | < 2 ⟺ - 2 < f (x) + 4 < 2.

Subtracting 4 from all parts yields:

- 6 < f (x) < - 2.

Thus, for all $x$ satisfying $0 < | x - 2 | < δ$ , we have $f (x) < - 2 < 0$ , which proves that $f (x)$ is strictly negative in a deleted neighborhood of $x = 2$ .

Exercise 5.

Suppose $f (x) \leq g (x)$ for all $x \neq a$ in an open interval containing a. Assume $lim_{x \to a} f (x) = L$ and $lim_{x \to a} g (x) = M$ . Prove that $L \leq M$ .

Answer

$L \leq M$ .

Solution

Define $h (x) = g (x) - f (x)$ . Since $f (x) \leq g (x)$ , we have $h (x) \geq 0$ for all $x \neq a$ near a.
By the Limit Laws, $lim_{x \to a} h (x) = lim_{x \to a} g (x) - lim_{x \to a} f (x) = M - L$ .

Assume for contradiction that $L > M$ , so $M - L < 0$ .
By the Sign Preservation Theorem applied to $h (x)$ , since $lim_{x \to a} h (x) = M - L < 0$ , there exists an open interval $I$ containing a such that $h (x)$ has the same sign as $M - L$ for all $x \in I ∖ {a}$ ; that is, $h (x) < 0$ .

However, $h (x) < 0$ contradicts $h (x) \geq 0$ . Therefore, our assumption $L > M$ must be false, which proves $L \leq M$ .

Exercise 6.

Prove that if $f (x) < 0$ for all $x \neq a$ in an open interval containing a, and if $lim_{x \to a} f (x) = L$ exists, then $L$ cannot be positive; that is, $L \leq 0$ .

Answer

$L \leq 0$ .

Solution

Suppose for contradiction that $lim_{x \to a} f (x) = L > 0$ .
By the Sign Preservation Theorem, since $L > 0$ , there exists an open interval $I_{1}$ containing a such that $f (x) > 0$ for all $x \in I_{1} ∖ {a}$ .

However, we are given that $f (x) < 0$ for all $x \neq a$ in some open interval $I_{2}$ containing a.

Taking the intersection $I = I_{1} \cap I_{2}$ , which is a non-empty open interval containing a, we find that for any $x \in I ∖ {a}$ , $f (x)$ must be simultaneously positive ( $f (x) > 0$ ) and negative ( $f (x) < 0$ ), which is impossible.

Thus, the assumption $L > 0$ leads to a contradiction. We conclude that $lim_{x \to a} f (x) = L \leq 0$ .

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