Function Limits and Continuity

Section 1: ε-δ Definition of Function Limits

Definition 1.1: Limit of a Function at a Point (ε-δ Definition)

Let $f$ be defined on some punctured neighborhood $U'(x_0, \delta_0)$ . We say $\lim_{x \to x_0} f(x) = L$ if and only if:

\forall \varepsilon > 0, \exists \delta > 0: 0 < |x - x_0| < \delta \Rightarrow |f(x) - L| < \varepsilon

Theorem 1.1: Uniqueness of Function Limits

If $\lim_{x \to x_0} f(x) = L_1$ and $\lim_{x \to x_0} f(x) = L_2$ , then $L_1 = L_2$ .

Proof of Theorem 1.1:

Suppose $L_1 \neq L_2$ . Let $\varepsilon = |L_1 - L_2|/2 > 0$ .

By the limit definitions, $\exists \delta_1, \delta_2 > 0$ such that for $0 < |x - x_0| < \delta_i$ : $|f(x) - L_i| < \varepsilon$ .

Let $\delta = \min(\delta_1, \delta_2)$ . Take any x with $0 < |x - x_0| < \delta$ .

By triangle inequality: $|L_1 - L_2| \leq |L_1 - f(x)| + |f(x) - L_2| < 2\varepsilon = |L_1 - L_2|$

Contradiction! Therefore $L_1 = L_2$ . ∎

∎

Example 1.1: Linear Function

Prove: $\lim_{x \to 2} (3x - 1) = 5$

Proof:

Given $\varepsilon > 0$ , choose $\delta = \varepsilon/3$ .

If $0 < |x - 2| < \delta$ , then:

|3x - 1 - 5| = |3x - 6| = 3|x - 2| < 3\delta = \varepsilon

Therefore the limit is 5. ∎

Example 1.2: Quadratic Function

Prove: $\lim_{x \to 3} x^2 = 9$

Proof:

We have $|x^2 - 9| = |x-3||x+3|$ .

If $|x - 3| < 1$ , then $2 < x < 4$ , so $|x+3| < 7$ .

Given $\varepsilon > 0$ , choose $\delta = \min(1, \varepsilon/7)$ .

If $0 < |x - 3| < \delta$ , then:

|x^2 - 9| = |x-3||x+3| < \delta \cdot 7 \leq \varepsilon

Therefore the limit is 9. ∎

Section 2: Heine's Theorem

Theorem 2.1: Heine's Theorem (Sequential Criterion)

Let f be defined on U'(x₀, δ₀). Then:

\lim_{x \to x_0} f(x) = L \quad \Longleftrightarrow \quad \text{For all } \{x_n\} \text{ with } x_n \to x_0, x_n \neq x_0: f(x_n) \to L

Example 2.1: Using Heine's Theorem

Show: $\lim_{x \to 0} \sin(1/x)$ does not exist.

Proof:

Consider sequences $x_n = 1/(2\pi n)$ and $y_n = 1/(\pi/2 + 2\pi n)$ .

Both converge to 0, but $\sin(1/x_n) = 0$ and $\sin(1/y_n) = 1$ .

By Heine's theorem, the limit does not exist. ∎

Section 3: One-Sided Limits

Definition 3.1: Right-Hand Limit

We say $\lim_{x \to x_0^+} f(x) = L$ if:

\forall \varepsilon > 0, \exists \delta > 0: x_0 < x < x_0 + \delta \Rightarrow |f(x) - L| < \varepsilon

Definition 3.2: Left-Hand Limit

We say $\lim_{x \to x_0^-} f(x) = L$ if:

\forall \varepsilon > 0, \exists \delta > 0: x_0 - \delta < x < x_0 \Rightarrow |f(x) - L| < \varepsilon

Theorem 3.1: Two-Sided vs One-Sided Limits

\lim_{x \to x_0} f(x) = L \quad \Longleftrightarrow \quad \lim_{x \to x_0^+} f(x) = L = \lim_{x \to x_0^-} f(x)

Section 4: Continuity

Definition 4.1: Continuity at a Point

Let $f$ be defined on an interval $I$ containing $x_0$ . We say $f$ is continuous at $x_0$ if:

\lim_{x \to x_0} f(x) = f(x_0)

Equivalently: $\forall \varepsilon > 0, \exists \delta > 0: |x - x_0| < \delta \Rightarrow |f(x) - f(x_0)| < \varepsilon$

Example 4.1: Verifying Continuity

Prove: $f(x) = 3x + 2$ is continuous at $x_0 = 1$ .

Proof:

We have $f(1) = 5$ and $\lim_{x \to 1} (3x + 2) = 5$ .

Since the limit equals the function value, f is continuous at 1. ∎

Section 5: Discontinuity Classification

Definition 5.1: Removable Discontinuity

$\lim_{x \to x_0} f(x)$ exists, but ≠ $f(x_0)$ (or $f(x_0)$ undefined).

Definition 5.2: Jump Discontinuity

Both $\lim_{x \to x_0^+} f(x)$ and $\lim_{x \to x_0^-} f(x)$ exist, but are unequal.

Definition 5.3: Essential Discontinuity

At least one one-sided limit doesn't exist (infinite or oscillating).

Example 5.1: Removable Discontinuity

For $f(x) = \frac{x^2 - 1}{x - 1}$ :

$f$ is undefined at $x = 1$ , but $\lim_{x \to 1} \frac{(x-1)(x+1)}{x-1} = 2$ .

This is a removable discontinuity.

Section 6: Limit Properties and Operations

Theorem 6.1: Limit Laws

If $\lim_{x \to x_0} f(x) = L$ and $\lim_{x \to x_0} g(x) = M$ , then:

$\lim(f + g) = L + M$
$\lim(f - g) = L - M$
$\lim(f \cdot g) = L \cdot M$
$\lim(f/g) = L/M$ (if $M \neq 0$ )
$\lim(cf) = cL$ for any constant $c$

Theorem 6.2: Squeeze Theorem for Functions

If $f(x) \leq g(x) \leq h(x)$ near $x_0$ and:

\lim_{x \to x_0} f(x) = \lim_{x \to x_0} h(x) = L

Then $\lim_{x \to x_0} g(x) = L$ .

Example 6.1: Using Squeeze Theorem

Find: $\lim_{x \to 0} x \sin(1/x)$

Solution: Since $-1 \leq \sin(1/x) \leq 1$ :

-|x| \leq x \sin(1/x) \leq |x|

Since $-|x| \to 0$ and $|x| \to 0$ as $x \to 0$ , by Squeeze Theorem:

\lim_{x \to 0} x \sin(1/x) = 0

Section 7: Local Boundedness and Sign Preservation

Theorem 7.1: Local Boundedness

If $\lim_{x \to x_0} f(x) = L$ exists, then $f$ is bounded in some neighborhood of $x_0$ .

Proof of Theorem 7.1:

Take $\epsilon = 1$ . Then $\exists \delta > 0$ such that:

0 < |x - x_0| < \delta \Rightarrow |f(x) - L| < 1

Therefore $|f(x)| < |L| + 1$ for $x \in (x_0 - \delta, x_0 + \delta) \setminus \{x_0\}$ .

∎

Theorem 7.2: Sign Preservation

If $\lim_{x \to x_0} f(x) = L > 0$ , then there exists $\delta > 0$ such that $f(x) > 0$ for all $x$ with $0 < |x - x_0| < \delta$ .

Similarly, if $L < 0$ , then $f(x) < 0$ near $x_0$ .

Example 7.1: Application

If $\lim_{x \to 0} f(x) = 5$ , then for $x$ sufficiently close to 0, we have $f(x) > 0$ .

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Practice Quiz: Function Limits and Continuity

10

Questions

0

Correct

0%

Accuracy

1

In the ε-δ definition of

\lim_{x \to x_0} f(x) = L

, what must be true?

Medium

Not attempted

2

What does Heine's theorem state?

Medium

Not attempted

3

For

\lim_{x \to 0} \sin(1/x)

, what can we conclude?

Medium

Not attempted

4

What is

\lim_{x \to 1} \frac{\sqrt{x} - 1}{x - 1}

?

Easy

Not attempted

5

A function is continuous at x₀ if:

Easy

Not attempted

6

If left and right limits differ, the two-sided limit:

Easy

Not attempted

7

What is

\lim_{x \to 0^+} \frac{1}{x}

?

Easy

Not attempted

8

A removable discontinuity occurs when:

Medium

Not attempted

9

To prove

\lim_{x \to 2} (3x - 1) = 5

, choose

\delta =

Hard

Not attempted

10

What is

\lim_{x \to 0} \frac{\sin 5x}{x}

?

Medium

Not attempted

Frequently Asked Questions

Why is 0 < |x - x₀| < δ used instead of |x - x₀| < δ?

The condition excludes x = x₀. The limit describes behavior as x approaches x₀, not the value at x₀. The function may not even be defined at x₀.

What's the difference between ε-N and ε-δ definitions?

For sequences, N is discrete (natural number). For functions, δ measures continuous distance. Heine's theorem bridges them.

How do I choose δ when proving a limit?

Work backwards: start with |f(x) - L| < ε, manipulate to get |x - x₀| < something. That 'something' becomes your δ.

What are the three conditions for continuity?

f(x₀) is defined, lim_{x→x₀} f(x) exists, and lim_{x→x₀} f(x) = f(x₀). All three must hold.

How do one-sided and two-sided limits relate?

Two-sided limit exists iff both one-sided limits exist and are equal. Continuity requires the two-sided limit to equal f(x₀).