CALC-3.1

4-5 hours

Function Limits and the ε-δ Definition

Master the precise definition of function limits that forms the rigorous foundation of calculus.

Learning Objectives

State and apply the precise ε-δ definition of function limits
Understand the geometric and intuitive meaning of the definition
Use Heine's theorem to connect function limits with sequence limits
Define and compute one-sided limits (left and right limits)
Understand and evaluate limits at infinity
Work with infinite limits and identify vertical asymptotes
Prove limits rigorously using the ε-δ definition
Apply standard limit formulas to compute complex limits

Historical Context

The ε-δ definition was developed in the 19th century to put calculus on a rigorous foundation. Augustin-Louis Cauchy (1789-1857) first introduced limits as the foundation of calculus in 1821.

Karl Weierstrass (1815-1897) formalized this into the precise ε-δ language we use today, resolving centuries of philosophical debates about infinitesimals.

1. The ε-δ Definition of Function Limits

In Chapter 2, we studied limits of sequences using the ε-N definition. Now we extend this concept to functions. The key idea remains the same: we want to capture what it means for $f(x)$ to get "arbitrarily close" to a value $L$ as $x$ gets "sufficiently close" to $x_0$ .

Definition 3.1: Punctured Neighborhood

A punctured neighborhood of $x_0$ with radius $\delta > 0$ is:

U'(x_0, \delta) = \{x \in \mathbb{R} : 0 < |x - x_0| < \delta\} = (x_0 - \delta, x_0) \cup (x_0, x_0 + \delta)

This is an open interval centered at $x_0$ , but with $x_0$ itself removed.

Definition 3.2: 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:

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

Understanding the Definition

1. "∀ε > 0" - The Challenge

Think of ε as a tolerance. Someone says: "I want f(x) within ε of L." No matter how small ε is, you must meet this challenge.

2. "∃δ > 0" - Your Response

You must produce a δ that works. The δ can depend on ε—smaller ε typically requires smaller δ.

3. "0 < |x - x₀| < δ" - Near but Not At

x is within δ of x₀, but NOT equal to x₀. We examine points near x₀, not at x₀ itself.

4. "|f(x) - L| < ε" - The Goal

f(x) must be within ε of L. If you can always achieve this for any ε, the limit is L.

Remark 3.1: Critical Observations

f need not be defined at x₀: We only require f to be defined on U'(x₀, δ₀).
f(x₀) is irrelevant: The limit depends only on values f(x) for x ≠ x₀.
δ depends on ε: We often write δ = δ(ε) to emphasize this.
Order of quantifiers matters: "∀ε, ∃δ" means δ can depend on ε.

Theorem 3.1: 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 3.1:

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 $L_1 = L_2$ . ∎

∎

Strategy for ε-δ Proofs

Scratch Work: Start with |f(x) - L| < ε. Work backwards to find how δ relates to ε.
Choose δ: Based on analysis, write δ = ... (often δ = ε/c for some constant c).
Formal Proof: Let ε > 0. Define δ. Verify: 0 < |x - x₀| < δ ⟹ |f(x) - L| < ε.

Example 3.1: Linear Function: lim(3x - 1) as x → 2 = 5

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

Scratch Work:

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

This holds when $|x - 2| < \varepsilon/3$ . So choose $\delta = \varepsilon/3$ .

Formal Proof:

Given: ε > 0. Choose: δ = ε/3 > 0.

Verify: If 0 < |x - 2| < δ = ε/3, then:

|3x - 1 - 5| = 3|x - 2| < 3 \cdot \frac{\varepsilon}{3} = \varepsilon \quad \checkmark

Example 3.2: Quadratic Function: lim x² as x → 3 = 9

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

Scratch Work:

Need $|x^2 - 9| = |x-3||x+3| < \varepsilon$

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

Then $|x^2 - 9| < 7|x - 3|$ . Need $|x - 3| < \varepsilon/7$ .

Formal Proof:

Choose: $\delta = \min(1, \varepsilon/7)$

If 0 < |x - 3| < δ, then |x - 3| < 1 (so |x+3| < 7) and |x - 3| < ε/7.

|x^2 - 9| = |x-3||x+3| < \frac{\varepsilon}{7} \cdot 7 = \varepsilon \quad \checkmark

Example 3.3: Rationalization: lim (√x - 1)/(x - 1) as x → 1 = 1/2

Claim: $\lim_{x \to 1} \frac{\sqrt{x} - 1}{x - 1} = \frac{1}{2}$

Simplification:

Note: $x - 1 = (\sqrt{x} - 1)(\sqrt{x} + 1)$ . For x ≠ 1:

\frac{\sqrt{x} - 1}{x - 1} = \frac{1}{\sqrt{x} + 1}

Analysis:

Need $\left|\frac{1}{\sqrt{x}+1} - \frac{1}{2}\right| = \frac{|1-\sqrt{x}|}{2(\sqrt{x}+1)} < \varepsilon$

If x > 0, then √x + 1 > 1, and |1 - √x| ≤ |1 - x|.

So the expression is bounded by |x - 1|/2.

Conclusion:

Choose δ = min(1/2, 2ε). Then the limit equals 1/2. ∎

Example 3.4: Non-existence: lim sin(1/x) as x → 0

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

Proof by Contradiction:

Suppose $\lim_{x \to 0} \sin(1/x) = L$ . Take ε = 1/2.

For any δ > 0, consider $x_n = 1/(2\pi n)$ and $y_n = 1/(\pi/2 + 2\pi n)$ .

For large n, both are in (0, δ). But sin(1/xₙ) = 0 and sin(1/yₙ) = 1.

Both |0 - L| < 1/2 and |1 - L| < 1/2 cannot hold simultaneously.

Contradiction! The limit does not exist. ∎

Common Mistakes

Wrong quantifier order: "∃δ, ∀ε" instead of "∀ε, ∃δ"
Including x = x₀: Writing |x - x₀| < δ instead of 0 < |x - x₀| < δ
δ depending on x: Your δ must be fixed once ε is given
Confusing f(x₀) with limit: The limit has nothing to do with f(x₀)

2. Heine's Theorem (Sequential Criterion)

Heine's theorem bridges function limits (ε-δ) and sequence limits (ε-N). This allows us to use everything we learned about sequences to analyze functions.

Theorem 3.2: Heine's Theorem

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

Proof of Theorem 3.2:

(⇒) Function limit ⟹ Sequential limit:

Assume $\lim_{x \to x_0} f(x) = L$ . Let {xₙ} be any sequence with xₙ → x₀, xₙ ≠ x₀.

Given ε > 0, ∃δ > 0: 0 < |x - x₀| < δ ⟹ |f(x) - L| < ε.

Since xₙ → x₀, ∃N: n > N ⟹ |xₙ - x₀| < δ. Since xₙ ≠ x₀, we have 0 < |xₙ - x₀| < δ.

Therefore |f(xₙ) - L| < ε for n > N, proving f(xₙ) → L. ∎

(⇐) By contrapositive:

If the function limit ≠ L, then ∃ε₀ > 0 such that ∀δ > 0, ∃x with 0 < |x - x₀| < δ but |f(x) - L| ≥ ε₀.

Take δ = 1/n. Get xₙ with 0 < |xₙ - x₀| < 1/n but |f(xₙ) - L| ≥ ε₀.

Then xₙ → x₀ (since |xₙ - x₀| < 1/n → 0), xₙ ≠ x₀, but f(xₙ) ↛ L. ∎

∎

Corollary 3.1: Non-existence Criterion

If ∃ sequences {xₙ}, {yₙ} → x₀ (with terms ≠ x₀) such that $\lim f(x_n) \neq \lim f(y_n)$ , then $\lim_{x \to x_0} f(x)$ does not exist.

Example 3.5: Using Heine's Theorem

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

Sequence 1: xₙ = 1/(nπ) → 0. Then sin(nπ) = 0, so limit = 0.

Sequence 2: yₙ = 1/(π/2 + 2nπ) → 0. Then sin(π/2 + 2nπ) = 1, so limit = 1.

Since 0 ≠ 1, by Corollary 3.1, the limit does not exist. ∎

Remark 3.2: When to Use Heine's Theorem

To prove existence: Show ALL sequences give the same limit
To prove non-existence: Find TWO sequences giving different limits (easier!)
To transfer results: Apply sequence theorems to function limits

3. One-Sided Limits

Sometimes we want to consider the limit from only one direction—from the left or from the right. This is especially useful for piecewise functions or functions with jump discontinuities.

Definition 3.3: Right-Hand Limit

We say $\lim_{x \to x_0^+} f(x) = L$ (the right-hand limit) if:

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

We approach x₀ from the right (x > x₀).

Definition 3.4: Left-Hand Limit

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

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

We approach x₀ from the left (x < x₀).

Theorem 3.3: 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)

The two-sided limit exists iff both one-sided limits exist and are equal.

Example 3.6: Piecewise Function

For $f(x) = \begin{cases} x^2 & x < 1 \\ 2x & x \geq 1 \end{cases}$ , find limits at x = 1.

Left-hand limit: For x < 1, f(x) = x². So $\lim_{x \to 1^-} f(x) = 1^2 = 1$

Right-hand limit: For x ≥ 1, f(x) = 2x. So $\lim_{x \to 1^+} f(x) = 2(1) = 2$

Conclusion: Since 1 ≠ 2, $\lim_{x \to 1} f(x)$ does not exist.

Example 3.7: Sign Function

For sgn(x) = {1 if x > 0, 0 if x = 0, -1 if x < 0}, find limits at x = 0.

\lim_{x \to 0^+} \text{sgn}(x) = 1, \quad \lim_{x \to 0^-} \text{sgn}(x) = -1

Since 1 ≠ -1, $\lim_{x \to 0} \text{sgn}(x)$ does not exist.

Example 3.8: Absolute Value

For $f(x) = \frac{|x|}{x}$ , find limits at x = 0.

For x > 0: |x|/x = x/x = 1. For x < 0: |x|/x = -x/x = -1.

\lim_{x \to 0^+} \frac{|x|}{x} = 1, \quad \lim_{x \to 0^-} \frac{|x|}{x} = -1

The two-sided limit does not exist (this is the sign function!).

4. Limits at Infinity

We can also ask: what happens to f(x) as x becomes very large (positive or negative)? This leads to the concept of limits at infinity.

Definition 3.5: Limit as x → +∞

Let f be defined on (a, +∞) for some a ∈ ℝ. We say $\lim_{x \to +\infty} f(x) = L$ if:

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

Definition 3.6: Limit as x → −∞

Similarly, $\lim_{x \to -\infty} f(x) = L$ if:

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

Remark 3.3: Horizontal Asymptotes

If $\lim_{x \to +\infty} f(x) = L$ or $\lim_{x \to -\infty} f(x) = L$ , the line y = L is a horizontal asymptote.

Rational Function Limits at Infinity

For $f(x) = \frac{a_nx^n + \cdots + a_0}{b_mx^m + \cdots + b_0}$ with aₙ, bₘ ≠ 0:

If n < m

\lim_{x \to \infty} f(x) = 0

If n = m

\lim_{x \to \infty} f(x) = \frac{a_n}{b_m}

If n > m

\lim_{x \to \infty} f(x) = \pm\infty

Example 3.9: Rational Function at Infinity

Find $\lim_{x \to +\infty} \frac{3x^2 + 2x - 1}{x^2 - 2}$ .

Method: Divide numerator and denominator by x² (highest power):

\frac{3x^2 + 2x - 1}{x^2 - 2} = \frac{3 + \frac{2}{x} - \frac{1}{x^2}}{1 - \frac{2}{x^2}}

As x → +∞: 2/x → 0, 1/x² → 0, 2/x² → 0.

\lim_{x \to +\infty} \frac{3x^2 + 2x - 1}{x^2 - 2} = \frac{3 + 0 - 0}{1 - 0} = 3

Example 3.10: Exponential vs Polynomial

Find $\lim_{x \to +\infty} \frac{e^x}{x^{100}}$ .

Answer: +∞. Exponentials always dominate polynomials.

Intuition: eˣ grows faster than any polynomial xⁿ as x → +∞.

5. Infinite Limits and Asymptotes

Definition 3.7: Infinite Limit at a Point

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

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

Similarly for $-\infty$ : replace f(x) > M with f(x) < -M.

Remark 3.4: Vertical Asymptotes

If $\lim_{x \to x_0^+} f(x) = \pm\infty$ or $\lim_{x \to x_0^-} f(x) = \pm\infty$ , the line x = x₀ is a vertical asymptote.

Example 3.11: Vertical Asymptote of 1/x

Find one-sided limits of f(x) = 1/x at x = 0.

\lim_{x \to 0^+} \frac{1}{x} = +\infty, \quad \lim_{x \to 0^-} \frac{1}{x} = -\infty

The line x = 0 is a vertical asymptote of y = 1/x.

Example 3.12: Finding All Asymptotes

Find all asymptotes of $f(x) = \frac{x^2 + 1}{x - 1}$ .

Vertical: x = 1 (denominator = 0).

\lim_{x \to 1^+} f(x) = +\infty, \quad \lim_{x \to 1^-} f(x) = -\infty

Oblique: Perform polynomial division: f(x) = x + 1 + 2/(x-1).

As x → ±∞, 2/(x-1) → 0, so y = x + 1 is an oblique asymptote.

6. Important Standard Limits

Fundamental Limits to Memorize

Trigonometric

\lim_{x \to 0} \frac{\sin x}{x} = 1

\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}

\lim_{x \to 0} \frac{\tan x}{x} = 1

Exponential/Logarithmic

\lim_{x \to 0} (1 + x)^{1/x} = e

\lim_{x \to 0} \frac{e^x - 1}{x} = 1

\lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1

Power Functions

\lim_{x \to 0} \frac{(1+x)^\alpha - 1}{x} = \alpha

\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a

Inverse Trig

\lim_{x \to 0} \frac{\arcsin x}{x} = 1

\lim_{x \to 0} \frac{\arctan x}{x} = 1

Example 3.13: Using Standard Limits

Find $\lim_{x \to 0} \frac{\sin 3x}{x}$ .

\lim_{x \to 0} \frac{\sin 3x}{x} = \lim_{x \to 0} \frac{\sin 3x}{3x} \cdot 3 = 1 \cdot 3 = 3

Example 3.14: Exponential Limit

Find $\lim_{x \to 0} (1 - 2x)^{1/x}$ .

Method: Rewrite using the standard limit form:

(1 - 2x)^{1/x} = \left[(1 + (-2x))^{1/(-2x)}\right]^{-2}

As x → 0, -2x → 0, so (1 + (-2x))^{1/(-2x)} → e.

\lim_{x \to 0} (1 - 2x)^{1/x} = e^{-2}

Example 3.15: Combined Techniques

Find $\lim_{x \to 0} \frac{e^{2x} - 1 - 2x}{x^2}$ .

Using Taylor expansion: e^{2x} ≈ 1 + 2x + 2x² + ... near x = 0.

\frac{e^{2x} - 1 - 2x}{x^2} \approx \frac{2x^2 + O(x^3)}{x^2} \to 2

Answer: 2

Example 3.16: Trigonometric Substitution

Find $\lim_{x \to 0} \frac{1 - \cos x}{x^2}$ .

Method 1 (Half-angle identity):

1 - \cos x = 2\sin^2(x/2)

\frac{1 - \cos x}{x^2} = \frac{2\sin^2(x/2)}{x^2} = \frac{1}{2} \cdot \left(\frac{\sin(x/2)}{x/2}\right)^2

As x → 0, (x/2) → 0, so sin(x/2)/(x/2) → 1.

\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \cdot 1^2 = \frac{1}{2}

Example 3.17: Composite Functions

Find $\lim_{x \to 0} \frac{\sin(\tan x)}{x}$ .

Solution: Write as a product:

\frac{\sin(\tan x)}{x} = \frac{\sin(\tan x)}{\tan x} \cdot \frac{\tan x}{x}

As x → 0: tan x → 0, so sin(tan x)/(tan x) → 1.

Also, tan x / x = (sin x / x) · (1 / cos x) → 1 · 1 = 1.

\lim_{x \to 0} \frac{\sin(\tan x)}{x} = 1 \cdot 1 = 1

Example 3.18: Logarithmic Limit

Find $\lim_{x \to 0} \frac{\ln(1 + \sin x)}{x}$ .

Solution: Write as:

\frac{\ln(1 + \sin x)}{x} = \frac{\ln(1 + \sin x)}{\sin x} \cdot \frac{\sin x}{x}

As x → 0: sin x → 0, so ln(1 + sin x)/(sin x) → 1.

And sin x / x → 1.

\lim_{x \to 0} \frac{\ln(1 + \sin x)}{x} = 1 \cdot 1 = 1

7. Advanced ε-δ Proof Techniques

Mastering ε-δ proofs requires practice with various function types. Here we present systematic approaches for different situations.

Strategy 1: Linear Functions

For $f(x) = ax + b$ with $a \neq 0$ :

|f(x) - L| = |ax + b - L| = |a||x - x_0|

Choose: $\delta = \varepsilon / |a|$

Linear functions are the simplest—δ is directly proportional to ε.

Strategy 2: Polynomial Functions

For polynomials near x₀:

Factor: $|f(x) - L| = |x - x_0| \cdot g(x)|$ where g(x) is bounded near x₀
Restrict: Assume $|x - x_0| < 1$ first to bound g(x)
Find bound: $|g(x)| \leq M$ for some M
Choose: $\delta = \min(1, \varepsilon/M)$

Strategy 3: Rational Functions

For $f(x) = p(x)/q(x)$ where $q(x_0) \neq 0$ :

First bound the denominator away from 0
Use: $|q(x)| \geq |q(x_0)|/2$ when x is close to x₀
Combine with numerator analysis

Key insight: Control the denominator before controlling the whole expression.

Example 3.19: Detailed Polynomial Proof

Claim: $\lim_{x \to 2} x^3 = 8$

Step 1: Factor

|x^3 - 8| = |x - 2||x^2 + 2x + 4|

Step 2: Bound the second factor

If |x - 2| < 1, then 1 < x < 3, so:

|x^2 + 2x + 4| \leq 9 + 6 + 4 = 19

Step 3: Choose δ

Need $|x - 2| \cdot 19 < \varepsilon$ , so $|x - 2| < \varepsilon/19$ .

Choose: $\delta = \min(1, \varepsilon/19)$

Step 4: Verify

If 0 < |x - 2| < δ, then:

|x^3 - 8| = |x - 2||x^2 + 2x + 4| < \frac{\varepsilon}{19} \cdot 19 = \varepsilon \quad \checkmark

Example 3.20: Rational Function Proof

Claim: $\lim_{x \to 1} \frac{1}{x} = 1$

Step 1: Analyze

\left|\frac{1}{x} - 1\right| = \left|\frac{1 - x}{x}\right| = \frac{|x - 1|}{|x|}

Step 2: Bound the denominator

If |x - 1| < 1/2, then 1/2 < x < 3/2, so |x| > 1/2.

Thus 1/|x| < 2.

Step 3: Complete

\frac{|x - 1|}{|x|} < 2|x - 1|

Need 2|x - 1| < ε, so |x - 1| < ε/2.

Choose: δ = min(1/2, ε/2)

Remark 3.5: The min() Trick

When your proof requires two conditions on δ:

δ < a (to make some preliminary bound work)
δ < ε/c (to get the final estimate)

Choose $\delta = \min(a, \varepsilon/c)$ . This ensures BOTH conditions hold.

8. Proving Limits Do NOT Exist

Sometimes the limit doesn't exist. Understanding how to prove non-existence is just as important as proving existence.

Definition 3.8: Negation of the ε-δ Definition

$\lim_{x \to x_0} f(x) \neq L$ means:

\exists \varepsilon_0 > 0 \text{ such that } \forall \delta > 0, \exists x: 0 < |x - x_0| < \delta \text{ but } |f(x) - L| \geq \varepsilon_0

Note how the quantifiers are flipped: ∀ε becomes ∃ε₀, and ∃δ becomes ∀δ.

Three Ways to Prove Non-existence

Method 1: Heine's Theorem

Find two sequences xₙ → x₀ and yₙ → x₀ such that lim f(xₙ) ≠ lim f(yₙ).

Often the easiest method!

Method 2: One-sided Limits Differ

Show $\lim_{x \to x_0^+} f(x) \neq \lim_{x \to x_0^-} f(x)$ .

Method 3: Direct ε-δ Negation

Exhibit a specific ε₀ for which no δ works.

Example 3.21: Oscillation at a Point

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

Proof using Heine's theorem:

Sequence 1: xₙ = 1/(2πn) → 0. Then cos(2πn) = 1 → 1.

Sequence 2: yₙ = 1/(πn) → 0. Then cos(πn) = (-1)ⁿ, which doesn't converge.

Since f(yₙ) doesn't converge, the function limit cannot exist. ∎

Example 3.22: Jump Discontinuity

Claim: $\lim_{x \to 0} \lfloor x \rfloor$ does not exist (floor function).

Proof using one-sided limits:

For 0 < x < 1: ⌊x⌋ = 0, so $\lim_{x \to 0^+} \lfloor x \rfloor = 0$

For -1 < x < 0: ⌊x⌋ = -1, so $\lim_{x \to 0^-} \lfloor x \rfloor = -1$

Since 0 ≠ -1, the two-sided limit doesn't exist. ∎

Example 3.23: Unbounded Oscillation

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

Proof: Consider xₙ = 1/(π/2 + 2πn).

Then sin(1/xₙ) = 1, and:

\frac{1}{x_n}\sin(1/x_n) = (\pi/2 + 2\pi n) \cdot 1 = \frac{\pi}{2} + 2\pi n \to +\infty

The function is unbounded near 0, so no finite limit exists.

But also consider yₙ = 1/(3π/2 + 2πn), giving values → -∞.

So even +∞ or -∞ is not a valid limit. The limit simply doesn't exist. ∎

9. Additional Worked Examples

Example 3.24: L'Hôpital Preview

Find $\lim_{x \to 0} \frac{x - \sin x}{x^3}$ (without L'Hôpital).

Using Taylor series: sin x = x - x³/6 + x⁵/120 - ...

x - \sin x = x - (x - \frac{x^3}{6} + O(x^5)) = \frac{x^3}{6} + O(x^5)

\frac{x - \sin x}{x^3} = \frac{x^3/6 + O(x^5)}{x^3} = \frac{1}{6} + O(x^2) \to \frac{1}{6}

Example 3.25: Squeeze Theorem Application

Find $\lim_{x \to \infty} \frac{\sin x}{x}$ .

Solution: Since -1 ≤ sin x ≤ 1 for all x:

\frac{-1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x} \quad \text{for } x > 0

As x → +∞: -1/x → 0 and 1/x → 0.

By Squeeze Theorem: $\lim_{x \to \infty} \frac{\sin x}{x} = 0$

Example 3.26: Exponential Base

Find $\lim_{x \to 0^+} x^x$ .

Solution: Write $x^x = e^{x \ln x}$ .

First find $\lim_{x \to 0^+} x \ln x$ :

x \ln x = \frac{\ln x}{1/x}

As x → 0⁺: ln x → -∞ and 1/x → +∞. This is -∞/∞ form.

Let t = 1/x, so x = 1/t and x → 0⁺ means t → +∞:

x \ln x = \frac{-\ln t}{t} \to 0 \text{ as } t \to +\infty

Therefore: $\lim_{x \to 0^+} x^x = e^0 = 1$

Example 3.27: Nested Radicals

Find $\lim_{x \to 0} \frac{\sqrt{1+x} - \sqrt{1-x}}{x}$ .

Solution: Rationalize:

\frac{\sqrt{1+x} - \sqrt{1-x}}{x} \cdot \frac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}

= \frac{(1+x) - (1-x)}{x(\sqrt{1+x} + \sqrt{1-x})} = \frac{2x}{x(\sqrt{1+x} + \sqrt{1-x})}

= \frac{2}{\sqrt{1+x} + \sqrt{1-x}} \to \frac{2}{1+1} = 1

Example 3.28: Comparing Growth Rates

Prove that $\lim_{x \to +\infty} \frac{\ln x}{x^\alpha} = 0$ for any α > 0.

Proof: Let y = ln x, so x = eʸ and x → +∞ means y → +∞.

\frac{\ln x}{x^\alpha} = \frac{y}{e^{\alpha y}}

Since eᵅʸ grows faster than any polynomial in y:

\frac{y}{e^{\alpha y}} \to 0 \text{ as } y \to +\infty

Conclusion: Logarithms grow slower than ANY positive power. ∎

10. Growth Rate Hierarchy

Understanding how different functions grow as x → ∞ is essential for evaluating limits. Here's the complete hierarchy from slowest to fastest growth.

Growth Rate Ordering (Slowest to Fastest)

ln(ln x)Iterated logarithm

ln xLogarithm

x^α (α > 0)Power functions

a^x (a > 1)Exponential

x!Factorial

x^xTower function

Key insight: When computing limits of ratios, the faster-growing function "wins."

Example 3.29: Applying Growth Hierarchy

Evaluate these limits at +∞:

\lim_{x \to +\infty} \frac{x^{100}}{e^x} = 0

Exponential beats any polynomial.

\lim_{x \to +\infty} \frac{(\ln x)^{1000}}{x^{0.001}} = 0

Any positive power beats any power of logarithm.

\lim_{x \to +\infty} \frac{e^x}{x!} = 0

Factorial beats exponential.

Remark 3.6: Infinitesimals Near Zero

Near x = 0, the hierarchy reverses for "smallness":

$x^2$ is smaller than $x$ (faster approach to 0)
$e^{-1/x^2}$ is smaller than any $x^n$
$1 - \cos x \sim x^2/2$ (same order as $x^2$ )

11. Infinitesimals and Asymptotic Notation

Definition 3.9: Infinitesimal

A function $\alpha(x)$ is an infinitesimal as x → x₀ if:

\lim_{x \to x_0} \alpha(x) = 0

Definition 3.10: Order Comparison of Infinitesimals

Let α(x) and β(x) be infinitesimals as x → x₀ with β(x) ≠ 0 near x₀:

Same order: $\lim \frac{\alpha}{\beta} = c \neq 0$ (write α ~ cβ or α = O(β))
α higher order than β: $\lim \frac{\alpha}{\beta} = 0$ (write α = o(β))
α lower order than β: $\lim \frac{\alpha}{\beta} = \pm\infty$
Equivalent: $\lim \frac{\alpha}{\beta} = 1$ (write α ~ β)

Standard Equivalences as x → 0

\sin x \sim x

\tan x \sim x

\arcsin x \sim x

\arctan x \sim x

1 - \cos x \sim \frac{x^2}{2}

e^x - 1 \sim x

\ln(1+x) \sim x

(1+x)^\alpha - 1 \sim \alpha x

Usage: Replace equivalent infinitesimals in limits for easier computation.

Example 3.30: Using Equivalences

Find $\lim_{x \to 0} \frac{\tan x - \sin x}{x^3}$ .

Solution:

\tan x - \sin x = \frac{\sin x}{\cos x} - \sin x = \sin x \left(\frac{1 - \cos x}{\cos x}\right)

Using equivalences: sin x ~ x, 1 - cos x ~ x²/2, cos x → 1:

\tan x - \sin x \sim x \cdot \frac{x^2/2}{1} = \frac{x^3}{2}

\lim_{x \to 0} \frac{\tan x - \sin x}{x^3} = \lim_{x \to 0} \frac{x^3/2}{x^3} = \frac{1}{2}

Example 3.31: Careful with Equivalences

Find $\lim_{x \to 0} \frac{\sin x - x}{x^3}$ .

Warning: Cannot simply replace sin x ~ x here (would give 0/x³)!

Correct approach: Use Taylor series:

\sin x = x - \frac{x^3}{6} + O(x^5)

\frac{\sin x - x}{x^3} = \frac{-x^3/6 + O(x^5)}{x^3} = -\frac{1}{6} + O(x^2) \to -\frac{1}{6}

Lesson: Equivalences only work for products/quotients, not for sums where cancellation occurs.

Remark 3.7: When to Use Equivalences

Safe: Replacing factors in products or quotients
Unsafe: Replacing terms in sums/differences (cancellation may occur)
Alternative: When unsafe, use Taylor expansion instead

12. Writing Rigorous ε-δ Proofs

Complete ε-δ Proof Template

Claim: lim_{x → x₀} f(x) = L

Proof:

Let ε > 0 be given. [Start of proof]

Choose δ = ... [Your choice, may depend on ε]

Suppose 0 < |x - x₀| < δ. [Assume the hypothesis]

Then ... [Chain of inequalities]

Therefore |f(x) - L| < ε. [Reach the conclusion]

Since ε was arbitrary, lim_{x → x₀} f(x) = L. ∎

Example 3.32: Complete Formal Proof

Claim: $\lim_{x \to 4} \sqrt{x} = 2$

Scratch Work (not part of formal proof):

Need $|\sqrt{x} - 2| < \varepsilon$ .

|\sqrt{x} - 2| = \frac{|x - 4|}{\sqrt{x} + 2}

If |x - 4| < 2, then x > 2, so √x > √2, thus √x + 2 > 2.

So $|\sqrt{x} - 2| < \frac{|x - 4|}{2}$ . Need |x - 4|/2 < ε, i.e., |x - 4| < 2ε.

Formal Proof:

Proof: Let ε > 0 be given.

Choose δ = min(2, 2ε) > 0.

Suppose 0 < |x - 4| < δ.

Since δ ≤ 2, we have |x - 4| < 2, so x > 2 > 0, which gives √x + 2 > 2.

Then:

|\sqrt{x} - 2| = \frac{|x - 4|}{\sqrt{x} + 2} < \frac{|x - 4|}{2} < \frac{\delta}{2} \leq \frac{2\varepsilon}{2} = \varepsilon

Since ε was arbitrary, $\lim_{x \to 4} \sqrt{x} = 2$ . ∎

Remark 3.8: Proof Writing Tips

Always state "Let ε > 0" at the beginning
Clearly specify your choice of δ before using it
Verify δ > 0 (especially for min() constructions)
Show each step clearly in the inequality chain
End with the conclusion and ∎ or QED

13. Preview: Connection to Continuity

The concept of limits naturally leads to continuity—one of the most important properties in analysis. Here's a preview of how they connect.

Definition 3.11: Continuity (Preview)

A function f is continuous at x₀ if:

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

This requires three things: (1) f(x₀) exists, (2) the limit exists, (3) they're equal.

Remark 3.9: Limit vs Continuity

Limit: Only cares about behavior near x₀, not at x₀
Continuity: Requires limit to match the actual value at x₀
Example: f(x) = sin(x)/x has a limit at 0, but isn't continuous there (undefined)
Removable discontinuity: Limit exists but ≠ f(x₀) (can be "fixed")

Example 3.33: Making a Function Continuous

Define f(0) to make $f(x) = \frac{\sin x}{x}$ continuous at x = 0.

Solution: We need f(0) = lim_{x→0} (sin x)/x = 1.

Define:

f(x) = \begin{cases} \frac{\sin x}{x} & x \neq 0 \\ 1 & x = 0 \end{cases}

This function is continuous everywhere.

Example 3.34: Limit at Discontinuity

For $f(x) = \lfloor x \rfloor$ (floor function), find limits at x = 2.

Left-hand limit: For 1 < x < 2, ⌊x⌋ = 1.

\lim_{x \to 2^-} \lfloor x \rfloor = 1

Right-hand limit: For 2 ≤ x < 3, ⌊x⌋ = 2.

\lim_{x \to 2^+} \lfloor x \rfloor = 2

Note: f(2) = 2, so f is right-continuous at 2 but not left-continuous.

Example 3.35: Trigonometric at Infinity

Find $\lim_{x \to +\infty} \left(1 + \frac{1}{x}\right)^{x^2}$ .

Solution: Write as exponential:

\left(1 + \frac{1}{x}\right)^{x^2} = e^{x^2 \ln\left(1 + \frac{1}{x}\right)}

Using ln(1 + u) ≈ u - u²/2 + ... for small u = 1/x:

x^2 \ln\left(1 + \frac{1}{x}\right) \approx x^2 \left(\frac{1}{x} - \frac{1}{2x^2}\right) = x - \frac{1}{2} \to +\infty

Therefore the limit is $+\infty$ .

Remark 3.10: Key Takeaways for Exam Preparation

Master the ε-δ template: Practice writing formal proofs for linear, quadratic, and rational functions
Know Heine's theorem: Use sequences to prove non-existence of limits
Memorize standard limits: They appear frequently in problems
Recognize indeterminate forms: 0/0, ∞/∞, 0·∞, 1^∞, 0^0, ∞^0, ∞-∞
Practice equivalence substitution: Know when it's safe and when it's not

What's Next?

Now that you've mastered the definition of function limits, the next section covers:

Uniqueness of limits - rigorous proof
Local boundedness - limits imply bounded behavior
Sign preservation - positive limits give positive values
Limit laws - arithmetic operations on limits
Squeeze theorem - powerful tool for evaluating limits
Cauchy criterion - alternative existence condition

Chapter 3.1 Comprehensive Summary

Key Definitions

• ε-δ definition of function limits
• One-sided limits (left and right)
• Limits at infinity (x → ±∞)
• Infinite limits (f(x) → ±∞)

Key Theorems

• Uniqueness of limits
• Heine's theorem (sequential criterion)
• Two-sided = both one-sided

Asymptotes

• Vertical: infinite limit at x₀
• Horizontal: limit at ±∞
• Oblique: linear at ±∞

Standard Limits

• (sin x)/x → 1
• (1 + x)^{1/x} → e
• (eˣ - 1)/x → 1
• ln(1+x)/x → 1

Practice Quiz: ε-δ Definition

Questions

Correct

Accuracy

In the ε-δ definition of

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

, what must be true?

Medium

Not attempted

What does Heine's theorem state?

Medium

Not attempted

For

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

, what can we conclude?

Medium

Not attempted

What is

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

Easy

Not attempted

What is

\lim_{x \to +\infty} \frac{3x^2 + 2x - 1}{x^2 - 2}

Easy

Not attempted

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

Easy

Not attempted

What is

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

Easy

Not attempted

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

, what replaces the δ condition?

Medium

Not attempted

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

, the graph has:

Easy

Not attempted

To prove

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

, choose

\delta =

Hard

Not attempted

What is

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

Medium

Not attempted

What is

\lim_{x \to 0} (1 + 3x)^{1/x}

Hard

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.

Can a function have a limit where it's undefined?

Yes! The limit only concerns behavior approaching x₀. Example: lim(sin x)/x = 1 as x→0.

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

Two-sided limit exists iff both one-sided limits exist and are equal.

Why do we need rigorous ε-δ proofs?

Intuition can be misleading. ε-δ provides a precise criterion for proving theorems and resolving debates.

Study Tips for Success

For Understanding

Draw ε-δ diagrams for each example
Verbalize the definition: "For any challenge ε, I can respond with δ"
Compare with the sequence definition (ε-N)
Understand WHY x₀ is excluded (0 < |x - x₀|)

For Problem Solving

Always do scratch work first—find δ before writing the proof
For products, use min(δ₁, δ₂) when multiple conditions needed
Practice standard limit formulas until automatic
Use Heine's theorem to disprove limits with counterexample sequences

Quick Reference: Standard Limits

Trigonometric

$\lim_{x \to 0} \frac{\sin x}{x} = 1$

$\lim_{x \to 0} \frac{\tan x}{x} = 1$

$\lim_{x \to 0} \frac{1-\cos x}{x^2} = \frac{1}{2}$

$\lim_{x \to 0} \frac{\arcsin x}{x} = 1$

$\lim_{x \to 0} \frac{\arctan x}{x} = 1$

Exponential

$\lim_{x \to 0} (1+x)^{1/x} = e$

$\lim_{x \to \infty} (1+\frac{1}{x})^x = e$

$\lim_{x \to 0} \frac{e^x-1}{x} = 1$

$\lim_{x \to 0} \frac{a^x-1}{x} = \ln a$

$\lim_{x \to 0^+} x^x = 1$

Logarithmic

$\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1$

$\lim_{x \to 0} \frac{(1+x)^\alpha - 1}{x} = \alpha$

$\lim_{x \to \infty} \frac{\ln x}{x^\alpha} = 0$

$\lim_{x \to 0^+} x \ln x = 0$

$\lim_{x \to \infty} \frac{x^n}{e^x} = 0$

Essential Formulas to Memorize

The ε-δ Definition

\lim_{x \to x_0} f(x) = L \Leftrightarrow \forall \varepsilon > 0, \exists \delta > 0:

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

Heine's Theorem

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

For all {xₙ} → x₀ (xₙ ≠ x₀): f(xₙ) → L