CALC-3.3

4-5 hours

Continuity and Discontinuity

Understand when functions behave "nicely" and classify when they don't.

Learning Objectives

Define continuity at a point using limits
Understand left and right continuity
Classify types of discontinuities
Prove continuity of elementary functions
Apply operations preserving continuity
Analyze continuity of composite functions

1. Definition of Continuity

Definition 3.9: 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: $\forall \varepsilon > 0, \exists \delta > 0: |x - x_0| < \delta \Rightarrow |f(x) - f(x_0)| < \varepsilon$

Remark 3.7: Three Conditions for Continuity

$f(x_0)$ is defined
$\lim_{x \to x_0} f(x)$ exists
$\lim_{x \to x_0} f(x) = f(x_0)$

Definition 3.10: One-Sided Continuity

Right-continuous: $\lim_{x \to x_0^+} f(x) = f(x_0)$
Left-continuous: $\lim_{x \to x_0^-} f(x) = f(x_0)$

$f$ is continuous at $x_0$ iff it's both left and right continuous there.

Definition 3.11: Continuity on an Interval

$f \in C(I)$ means $f$ is continuous at every point of interval $I$ . At endpoints, we require one-sided continuity.

Example 3.10: Verifying Continuity with ε-δ

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

Proof: We need: $\forall \varepsilon > 0, \exists \delta > 0: |x - 1| < \delta \Rightarrow |f(x) - f(1)| < \varepsilon$

Scratch work:

|f(x) - f(1)| = |3x + 2 - 5| = |3x - 3| = 3|x - 1|

We want 3|x - 1| < ε, so |x - 1| < ε/3.

Formal proof: Given ε > 0, let δ = ε/3. Then:

|x - 1| < \delta \Rightarrow |f(x) - f(1)| = 3|x - 1| < 3\delta = \varepsilon \quad \checkmark

Example 3.10b: Continuity of Quadratic

Prove $f(x) = x^2$ is continuous at $x_0 = 2$ .

Scratch work:

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

Near x = 2, say |x - 2| < 1, we have 1 < x < 3, so |x + 2| < 5.

Formal proof: Given ε > 0, let δ = min(1, ε/5). For |x - 2| < δ:

|x^2 - 4| = |x - 2||x + 2| < \delta \cdot 5 \leq \varepsilon \quad \checkmark

Remark 3.7b: Intuition: No Jumps, No Gaps

A function is continuous at x₀ if:

No gap: f(x₀) is defined
No jump: f(x) approaches f(x₀) as x → x₀
No disagreement: the limit equals the function value

Graphically: you can draw the function without lifting your pen.

Sequential Characterization

$f$ is continuous at $x_0$ if and only if:

\forall \{x_n\} \to x_0: \lim_{n \to \infty} f(x_n) = f(x_0)

This is often easier to use for proving discontinuity: find a sequence approaching x₀ where f(xₙ) doesn't approach f(x₀).

2. Classification of Discontinuities

Types of Discontinuity

Removable Discontinuity

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

Example: $f(x) = \frac{\sin x}{x}$ at $x = 0$

Jump Discontinuity (First Kind)

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

Example: $\text{sgn}(x)$ at $x = 0$

Essential Discontinuity (Second Kind)

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

Example: $\sin(1/x)$ at $x = 0$ , or $1/x$ at $x = 0$

Example 3.11: Removable Discontinuity

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

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

Define $\tilde{f}(1) = 2$ to "remove" the discontinuity.

Example 3.12: Jump Discontinuity

The Heaviside step function $H(x) = \begin{cases} 0 & x < 0 \\ 1 & x \geq 0 \end{cases}$ :

$\lim_{x \to 0^-} H(x) = 0$ , $\lim_{x \to 0^+} H(x) = 1$

Jump of magnitude $|1 - 0| = 1$ .

Example 3.13: Essential Discontinuity

$f(x) = \sin(1/x)$ at $x = 0$ :

Neither $\lim_{x \to 0^+}$ nor $\lim_{x \to 0^-}$ exists (infinite oscillation).

Example 3.13b: Infinite Discontinuity

$f(x) = 1/x$ at $x = 0$ :

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

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

Both limits are infinite, so this is an essential (second kind) discontinuity.

Example 3.13c: Mixed Example: Piecewise Function

Classify discontinuities of:

f(x) = \begin{cases} x^2 & x < 0 \\ 1 & x = 0 \\ 2x & x > 0 \end{cases}

At x = 0:

$\lim_{x \to 0^-} x^2 = 0$
$\lim_{x \to 0^+} 2x = 0$
Both one-sided limits equal 0, so $\lim_{x \to 0} f(x) = 0$
But f(0) = 1 ≠ 0

Conclusion: Removable discontinuity at x = 0.

Example 3.13d: Dirichlet Function

The Dirichlet function:

D(x) = \begin{cases} 1 & x \in \mathbb{Q} \\ 0 & x \notin \mathbb{Q} \end{cases}

Claim: D(x) is discontinuous at every point.

Proof: For any x₀ and any neighborhood, there exist both rationals and irrationals arbitrarily close to x₀. So D takes values 0 and 1 arbitrarily close to x₀, meaning no limit exists.

Every point is an essential discontinuity!

Key Distinction

First Kind (Removable or Jump): Both one-sided limits exist (and are finite).

Second Kind (Essential): At least one one-sided limit is infinite or doesn't exist.

First kind discontinuities are "mild"—the function has a definite behavior on each side. Second kind discontinuities are "wild."

Theorem 3.11b: Jump Size

At a jump discontinuity x₀, the jump is defined as:

\sigma(x_0) = \lim_{x \to x_0^+} f(x) - \lim_{x \to x_0^-} f(x)

The sign indicates direction: positive = jump up, negative = jump down.

3. Operations on Continuous Functions

Theorem 3.12: Algebraic Operations

If $f, g$ are continuous at $x_0$ , then so are:

$f + g$ , $f - g$
$f \cdot g$
$f/g$ (provided $g(x_0) \neq 0$ )
$|f|$ , $\max(f, g)$ , $\min(f, g)$

Theorem 3.13: Composition of Continuous Functions

If $f$ is continuous at $x_0$ and $g$ is continuous at $f(x_0)$ , then $g \circ f$ is continuous at $x_0$ :

\lim_{x \to x_0} g(f(x)) = g(f(x_0)) = g\left(\lim_{x \to x_0} f(x)\right)

Remark 3.8: Limit Can Pass Inside Continuous Functions

For continuous $g$ : $\lim g(f(x)) = g(\lim f(x))$ . Very useful!

Proof:

Proof of Sum Rule:

Let f, g be continuous at x₀. We show f + g is continuous at x₀.

Given ε > 0. Since f is continuous at x₀, ∃δ₁ > 0: |x - x₀| < δ₁ ⟹ |f(x) - f(x₀)| < ε/2.

Since g is continuous at x₀, ∃δ₂ > 0: |x - x₀| < δ₂ ⟹ |g(x) - g(x₀)| < ε/2.

Let δ = min(δ₁, δ₂). For |x - x₀| < δ:

|(f+g)(x) - (f+g)(x_0)| \leq |f(x) - f(x_0)| + |g(x) - g(x_0)| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon

Thus f + g is continuous at x₀. ∎

∎

Proof:

Proof of Product Rule:

Let f, g be continuous at x₀. We show fg is continuous at x₀.

Key trick: Add and subtract f(x)g(x₀):

f(x)g(x) - f(x_0)g(x_0) = f(x)[g(x) - g(x_0)] + g(x_0)[f(x) - f(x_0)]

Since f is continuous at x₀, f is locally bounded: ∃M > 0, ∃δ₀ > 0: |x - x₀| < δ₀ ⟹ |f(x)| ≤ M.

Given ε > 0:

∃δ₁: |x - x₀| < δ₁ ⟹ |g(x) - g(x₀)| < ε/(2M)
∃δ₂: |x - x₀| < δ₂ ⟹ |f(x) - f(x₀)| < ε/(2|g(x₀)| + 1)

Let δ = min(δ₀, δ₁, δ₂). The result follows. ∎

∎

Example 3.14b: Combining Operations

Show $f(x) = \frac{x^2 \sin x}{x + 1}$ is continuous at x = 1.

Solution:

x² is continuous (polynomial)
sin x is continuous (elementary)
x + 1 is continuous and ≠ 0 at x = 1

By the product and quotient rules, f is continuous at x = 1.

Indeed: $f(1) = \frac{1 \cdot \sin 1}{2} = \frac{\sin 1}{2}$

Warning: Division by Zero

The quotient f/g is continuous only where g ≠ 0.

Example: $h(x) = \frac{x}{x-1}$ is continuous on (-∞, 1) ∪ (1, ∞), but not at x = 1.

Always check the denominator before claiming a quotient is continuous!

Theorem 3.13b: Composition Theorem (Detailed)

If:

$\lim_{x \to x_0} g(x) = L$
f is continuous at L

Then: $\lim_{x \to x_0} f(g(x)) = f(L) = f\left(\lim_{x \to x_0} g(x)\right)$

Proof:

Proof of Composition Theorem:

Given ε > 0. Since f is continuous at L:

\exists \eta > 0: |y - L| < \eta \Rightarrow |f(y) - f(L)| < \varepsilon

Since g(x) → L as x → x₀:

\exists \delta > 0: 0 < |x - x_0| < \delta \Rightarrow |g(x) - L| < \eta

Combining: 0 < |x - x₀| < δ ⟹ |g(x) - L| < η ⟹ |f(g(x)) - f(L)| < ε. ∎

∎

Example 3.14c: Using Composition

Find $\lim_{x \to 0} e^{\sin x}$ .

Solution:

$\lim_{x \to 0} \sin x = 0$
eˣ is continuous at 0

By composition: $\lim_{x \to 0} e^{\sin x} = e^{\lim_{x \to 0} \sin x} = e^0 = 1$

4. Continuity of Elementary Functions

Theorem 3.14: Elementary Functions are Continuous

The following are continuous on their natural domains:

Polynomials: continuous on $\mathbb{R}$
Rational functions $P(x)/Q(x)$ : continuous where $Q(x) \neq 0$
Exponential $a^x$ : continuous on $\mathbb{R}$
Logarithm $\log_a x$ : continuous on $(0, \infty)$
Trigonometric: $\sin, \cos$ on $\mathbb{R}$ ; $\tan$ where defined
Inverse trig: $\arcsin, \arccos, \arctan$ on their domains

Example 3.14: Continuity of Exponential

Claim: $a^x$ ( $a > 0$ ) is continuous at every $x_0 \in \mathbb{R}$ .

Proof sketch:

\lim_{x \to x_0} a^x = \lim_{h \to 0} a^{x_0 + h} = a^{x_0} \cdot \lim_{h \to 0} a^h = a^{x_0} \cdot 1 = a^{x_0}

Definition 3.12: Logarithm Function

For $a > 0, a \neq 1$ , $\log_a: (0, \infty) \to \mathbb{R}$ is the inverse of $a^x$ .

Natural log: $\ln x = \log_e x$ .

Properties of Logarithm

$\log_a(xy) = \log_a x + \log_a y$
$\log_a(x^\alpha) = \alpha \log_a x$
$\log_a a = 1$
$a^{\log_a x} = x$ , $\log_a(a^x) = x$

Example 3.15: Continuity of sin x

Prove sin x is continuous at every x₀ ∈ ℝ.

Proof: Using the identity:

\sin x - \sin x_0 = 2\cos\left(\frac{x + x_0}{2}\right)\sin\left(\frac{x - x_0}{2}\right)

Since |cos(·)| ≤ 1 and |sin(θ)| ≤ |θ|:

|\sin x - \sin x_0| \leq 2 \cdot 1 \cdot \left|\frac{x - x_0}{2}\right| = |x - x_0|

Given ε > 0, let δ = ε. Then |x - x₀| < δ ⟹ |sin x - sin x₀| < ε. ∎

Example 3.15b: Continuity of Inverse Functions

If f is continuous and strictly monotone on interval I, then f⁻¹ is continuous on f(I).

Application: Since eˣ is continuous and strictly increasing on ℝ, ln x is continuous on (0, ∞).

Similarly, since sin x is strictly increasing on [-π/2, π/2], arcsin x is continuous on [-1, 1].

5. Worked Examples

Example 3.16: Finding Points of Continuity

Find all points where $f(x) = \frac{x^2 - 4}{x^2 - 5x + 6}$ is continuous.

Solution: Factor:

f(x) = \frac{(x-2)(x+2)}{(x-2)(x-3)}

Denominator = 0 when x = 2 or x = 3.

At x = 2: $\lim_{x \to 2} \frac{x+2}{x-3} = \frac{4}{-1} = -4$

Limit exists but f(2) undefined → Removable discontinuity

At x = 3: $\lim_{x \to 3} \frac{x+2}{x-3}$ DNE (infinite)

→ Essential discontinuity

Continuous on: (-∞, 2) ∪ (2, 3) ∪ (3, ∞)

Example 3.17: Making a Function Continuous

Find c so that $f(x) = \begin{cases} x^2 + c & x < 1 \\ 2x + 1 & x \geq 1 \end{cases}$ is continuous.

Solution: For continuity at x = 1:

\lim_{x \to 1^-} (x^2 + c) = \lim_{x \to 1^+} (2x + 1)

1 + c = 3 \Rightarrow c = 2

Example 3.18: Analyzing Piecewise Continuity

Analyze continuity of:

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

At x = 0:

g(0) = 1 ✓
$\lim_{x \to 0} \frac{\sin x}{x} = 1$ ✓
Limit = g(0) ✓

→ Continuous at x = 0!

At x ≠ 0: sin x/x is a ratio of continuous functions with non-zero denominator → continuous.

Conclusion: g is continuous on all of ℝ.

Example 3.19: Thomae's Function

The "popcorn function":

T(x) = \begin{cases} 1/q & x = p/q \text{ (reduced)} \\ 0 & x \notin \mathbb{Q} \end{cases}

Remarkable fact:

T is continuous at every irrational point
T is discontinuous at every rational point

Why? Near any x₀, rationals p/q with large q have T(p/q) = 1/q ≈ 0 ≈ T(x₀) if x₀ is irrational.

Example 3.20: Continuity via Squeeze

Show $f(x) = x^2 \sin(1/x)$ with f(0) = 0 is continuous at 0.

Solution: For x ≠ 0:

|f(x) - f(0)| = |x^2 \sin(1/x)| \leq x^2 \cdot 1 = x^2

Given ε > 0, let δ = √ε. Then |x| < δ ⟹ |f(x)| ≤ x² < ε. ∎

6. Common Mistakes

Mistake 1: Forgetting to Check f(x₀)

Wrong: "lim f(x) = 3 as x → 2, so f is continuous at 2."

Right: Must also verify f(2) = 3. If f(2) is undefined or ≠ 3, there's a discontinuity.

Mistake 2: Assuming Quotients are Continuous

Wrong: "f/g is continuous because f and g are."

Right: f/g is continuous only where g ≠ 0.

Mistake 3: Confusing Discontinuity Types

Wrong: "The limit is infinite, so it's a jump discontinuity."

Right: Infinite limits indicate essential (second kind) discontinuity. Jump requires both one-sided limits to be finite.

Mistake 4: Wrong Domain Analysis

Wrong: "ln(x²) is continuous on ℝ because x² > 0."

Right: x² = 0 at x = 0, so ln(x²) is only continuous on (-∞, 0) ∪ (0, ∞).

7. Practice Problems

Problem Set A: Classification

Classify discontinuities of $f(x) = \frac{|x|}{x}$
(Answer: Jump at x = 0)
Classify discontinuities of $g(x) = \frac{x}{\sin x}$
(Answer: Removable at x = 0, essential at x = nπ, n ≠ 0)
Classify discontinuities of $h(x) = \lfloor x \rfloor$ (floor function)
(Answer: Jump at every integer)

Problem Set B: Finding Constants

Find a, b so $f(x) = \begin{cases} ax + b & x < 0 \\ \cos x & x \geq 0 \end{cases}$ is continuous and differentiable at x = 0.
(Answer: a = 0, b = 1)
Find c so $g(x) = \begin{cases} \frac{e^x - 1}{x} & x \neq 0 \\ c & x = 0 \end{cases}$ is continuous.
(Answer: c = 1)

Problem Set C: Proofs

Prove |f| is continuous at x₀ if f is continuous at x₀.
Prove max(f, g) is continuous if f and g are continuous.
Give an example where f² is continuous but f is not.

8. Additional Worked Examples

Example 3.21: Absolute Value Continuity

Prove |x| is continuous at every x₀ ∈ ℝ.

Proof: Using the reverse triangle inequality:

||x| - |x_0|| \leq |x - x_0|

Given ε > 0, let δ = ε. Then:

|x - x_0| < \delta \Rightarrow ||x| - |x_0|| \leq |x - x_0| < \varepsilon \quad \checkmark

Example 3.22: Square Root Continuity

Prove √x is continuous at every x₀ > 0.

Proof: For x, x₀ > 0:

|\sqrt{x} - \sqrt{x_0}| = \frac{|x - x_0|}{\sqrt{x} + \sqrt{x_0}} \leq \frac{|x - x_0|}{\sqrt{x_0}}

Given ε > 0, let δ = ε√x₀. Then |x - x₀| < δ implies:

|\sqrt{x} - \sqrt{x_0}| \leq \frac{\delta}{\sqrt{x_0}} = \varepsilon \quad \checkmark

Example 3.23: Continuity at x = 0 for √x

Show √x is right-continuous at x = 0.

Proof: Need: |√x - 0| < ε when 0 ≤ x < δ.

|√x| < ε ⟺ x < ε²

Let δ = ε². Then 0 ≤ x < δ ⟹ √x < √δ = ε. ∎

Example 3.24: Exponential Continuity

Prove eˣ is continuous at every x₀ ∈ ℝ.

Proof:

|e^x - e^{x_0}| = e^{x_0}|e^{x-x_0} - 1|

We use: for |h| < 1, |eʰ - 1| ≤ 2|h|.

Given ε > 0, let δ = min(1, ε/(2e^{x₀})). Then:

|e^x - e^{x_0}| = e^{x_0}|e^{x-x_0} - 1| \leq e^{x_0} \cdot 2|x - x_0| < e^{x_0} \cdot 2\delta \leq \varepsilon

Example 3.25: Rational Function Domain

Find all points of continuity for $f(x) = \frac{x^3 - 8}{x^2 - 4}$ .

Solution: Factor:

f(x) = \frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)}

At x = 2: $\lim_{x \to 2} \frac{x^2+2x+4}{x+2} = \frac{12}{4} = 3$

→ Removable discontinuity

At x = -2: Limit is ∞ → Essential discontinuity

Continuous on: ℝ \ {-2, 2}

Example 3.26: Trigonometric Continuity

Show tan x is continuous at x = 0.

Solution: tan x = sin x / cos x

sin x is continuous at 0
cos x is continuous at 0
cos(0) = 1 ≠ 0

By quotient rule, tan x is continuous at 0.

tan(0) = sin(0)/cos(0) = 0/1 = 0 ✓

Example 3.27: Composition Example

Show $h(x) = e^{\sin x}$ is continuous on ℝ.

Solution:

sin x is continuous on ℝ
eᵘ is continuous on ℝ
Composition of continuous functions is continuous

Therefore e^{sin x} is continuous on ℝ.

Example 3.28: Multiple Composition

Show $f(x) = \ln(1 + e^{\cos x})$ is continuous on ℝ.

Solution: Break down the composition:

cos x is continuous on ℝ
e^{cos x} is continuous (composition)
1 + e^{cos x} is continuous (sum)
Since e^{cos x} > 0, we have 1 + e^{cos x} > 1 > 0
ln(1 + e^{cos x}) is continuous (composition with ln)

Example 3.29: Piecewise with Three Pieces

Analyze continuity of:

f(x) = \begin{cases} x^2 & x < -1 \\ 2x + 3 & -1 \leq x < 2 \\ 7 - x & x \geq 2 \end{cases}

At x = -1:

Left limit: (-1)² = 1
Right limit: 2(-1) + 3 = 1
f(-1) = 2(-1) + 3 = 1

All equal → Continuous at x = -1

At x = 2:

Left limit: 2(2) + 3 = 7
Right limit: 7 - 2 = 5

Left ≠ Right → Jump discontinuity at x = 2

Example 3.30: Modified Dirichlet

Define:

f(x) = \begin{cases} x & x \in \mathbb{Q} \\ 0 & x \notin \mathbb{Q} \end{cases}

Claim: f is continuous only at x = 0.

At x = 0: For any sequence xₙ → 0, |f(xₙ)| ≤ |xₙ| → 0 = f(0). ✓

At x ≠ 0: Take rational sequence rₙ → x₀ and irrational sequence iₙ → x₀.

f(rₙ) = rₙ → x₀, but f(iₙ) = 0 → 0 ≠ x₀. Discontinuous.

9. Preview: Uniform Continuity

Pointwise vs Uniform Continuity

Pointwise Continuity (what we've studied)

∀x₀ ∀ε > 0 ∃δ > 0: |x - x₀| < δ ⟹ |f(x) - f(x₀)| < ε

δ can depend on both ε AND x₀.

Uniform Continuity (Chapter 3.4)

∀ε > 0 ∃δ > 0 ∀x₀: |x - x₀| < δ ⟹ |f(x) - f(x₀)| < ε

Same δ works for ALL x₀. Much stronger!

Remark 3.9: Key Theorem Preview

Heine-Cantor Theorem: Every continuous function on a closed bounded interval [a, b] is uniformly continuous.

This powerful result connects pointwise and uniform continuity on compact sets.

10. Study Tips and Strategies

Problem-Solving Strategy

Step 1: Identify the Type

• Is it a composition? Product? Quotient?
• Where might discontinuities occur?
• Is it piecewise?

Step 2: Check Critical Points

• Where denominator = 0
• Where pieces meet
• Domain boundaries

Step 3: Compute Limits

• One-sided limits at critical points
• Compare with function values
• Classify any discontinuities

Memory Aids

Discontinuity Types

• Removable = Repairable
• Jump = Just a step
• Essential = Extreme behavior

Three Conditions

1. f(x₀) Defined
2. Limit Exists
3. Same value
Mnemonic: DEStiny

11. More Practice Problems

Problem Set D: Advanced

Analyze continuity of $f(x) = \frac{\sin x - x}{x^3}$ at x = 0.
(Hint: Use Taylor expansion or L'Hôpital)
Show $g(x) = x \cdot D(x)$ where D is Dirichlet function is continuous only at x = 0.
Find all c such that $f(x) = \begin{cases} \frac{\ln(1+cx)}{x} & x \neq 0 \\ 3 & x = 0 \end{cases}$ is continuous.
(Answer: c = 3)
Prove: If f is continuous and f(x) = 0 for all rational x, then f ≡ 0.
Show $h(x) = \lim_{n \to \infty} \frac{x^{2n} - 1}{x^{2n} + 1}$ has jump discontinuities.

Problem Set E: True/False

If f is continuous at x₀, then |f| is continuous at x₀. (True)
If |f| is continuous at x₀, then f is continuous at x₀. (False)
If f² is continuous at x₀, then f is continuous at x₀. (False)
If f + g is continuous, both f and g are continuous. (False)
A function with only jump discontinuities is bounded on [a,b]. (True)

12. Detailed Proofs

Theorem 3.15: Continuous Functions Preserve Boundedness Locally

If f is continuous at x₀, then f is bounded on some neighborhood of x₀.

Proof:

Let ε = 1. Since f is continuous at x₀, ∃δ > 0:

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

This means |f(x)| < |f(x₀)| + 1 for all x ∈ (x₀ - δ, x₀ + δ).

So f is bounded by M = |f(x₀)| + 1 on this neighborhood. ∎

∎

Theorem 3.16: Absolute Value Preserves Continuity

If f is continuous at x₀, then |f| is continuous at x₀.

Proof:

By the reverse triangle inequality:

||f(x)| - |f(x_0)|| \leq |f(x) - f(x_0)|

Given ε > 0. Since f is continuous at x₀, ∃δ > 0: |x - x₀| < δ ⟹ |f(x) - f(x₀)| < ε.

Then: |x - x₀| < δ ⟹ ||f(x)| - |f(x₀)|| ≤ |f(x) - f(x₀)| < ε. ∎

∎

Theorem 3.17: Max and Min Preserve Continuity

If f, g are continuous at x₀, then max(f, g) and min(f, g) are continuous at x₀.

Proof:

Use the identities:

\max(f, g) = \frac{f + g + |f - g|}{2}

\min(f, g) = \frac{f + g - |f - g|}{2}

Since f, g are continuous, so is f - g (sum rule).

Since f - g is continuous, so is |f - g| (absolute value rule).

Since f + g and |f - g| are continuous, so is their sum and quotient by 2. ∎

∎

Example 3.31: Application of Max/Min

Show f(x) = max(x, x²) is continuous on ℝ.

Solution:

x is continuous (polynomial)
x² is continuous (polynomial)
By Theorem 3.17, max(x, x²) is continuous

Note: max(x, x²) = x² for x ≤ 0 or x ≥ 1, and = x for 0 < x < 1.

Theorem 3.18: Sign Preservation for Continuous Functions

If f is continuous at x₀ and f(x₀) > 0, then ∃δ > 0: f(x) > 0 for all x ∈ (x₀ - δ, x₀ + δ).

Proof:

Let ε = f(x₀)/2 > 0. Since f is continuous at x₀:

\exists \delta > 0: |x - x_0| < \delta \Rightarrow |f(x) - f(x_0)| < \frac{f(x_0)}{2}

This means:

f(x_0) - \frac{f(x_0)}{2} < f(x) < f(x_0) + \frac{f(x_0)}{2}

\frac{f(x_0)}{2} < f(x) < \frac{3f(x_0)}{2}

In particular, f(x) > f(x₀)/2 > 0. ∎

∎

Corollary 3.5: Sign Preservation (Negative)

If f is continuous at x₀ and f(x₀) < 0, then f(x) < 0 in some neighborhood of x₀.

13. Additional Examples

Example 3.32: Continuity with Nested Functions

Analyze continuity of $f(x) = \sqrt{|\sin x|}$ .

Solution:

sin x is continuous on ℝ
|sin x| is continuous (composition with |·|)
|sin x| ≥ 0, so √|sin x| is defined everywhere
√u is continuous on [0, ∞)
Composition: √|sin x| is continuous on ℝ

Example 3.33: Oscillatory Behavior

Compare continuity at x = 0 for:

f(x) = sin(1/x)
g(x) = x·sin(1/x)
h(x) = x²·sin(1/x)

f(x): lim DNE (oscillates) → Essential discontinuity at 0

g(x): |g(x)| ≤ |x| → 0. But need to define g(0) = 0. Then continuous.

h(x): |h(x)| ≤ x² → 0 even faster. With h(0) = 0, continuous.

Example 3.34: Fractional Exponents

Where is $f(x) = x^{1/3}$ continuous?

Solution: x^{1/3} (cube root) is defined for all x ∈ ℝ.

At x₀ ≠ 0: Use |a - b| = |a³ - b³|/|a² + ab + b²| with a = x^{1/3}, b = x₀^{1/3}.

At x₀ = 0: |x^{1/3}| = |x|^{1/3} → 0 as x → 0.

Conclusion: Continuous on all of ℝ.

Example 3.35: Complex Piecewise

Find all points of discontinuity:

f(x) = \begin{cases} \frac{x^2 - 1}{x - 1} & x < 1, x \neq -1 \\ 4 & x = 1 \\ 2^x & x > 1 \end{cases}

At x = -1: f is undefined (denominator = 0)

But lim_{x→-1} (x² - 1)/(x - 1) = lim (x + 1) = 0. → Removable discontinuity

At x = 1:

Left: lim_{x→1⁻} (x + 1) = 2
Right: lim_{x→1⁺} 2ˣ = 2
f(1) = 4 ≠ 2

→ Removable discontinuity at x = 1

Example 3.36: Proving Discontinuity via Sequences

Show f(x) = sin(1/x) is discontinuous at x = 0.

Proof: Consider two sequences approaching 0:

x_n = \frac{1}{n\pi} \to 0, \quad f(x_n) = \sin(n\pi) = 0

y_n = \frac{1}{\frac{\pi}{2} + 2n\pi} \to 0, \quad f(y_n) = \sin(\frac{\pi}{2} + 2n\pi) = 1

Different sequences give different limits (0 ≠ 1).

By sequential criterion, lim_{x→0} sin(1/x) DNE. ∎

14. Why Continuity Matters

Applications in Calculus

1. Intermediate Value Theorem

If f is continuous on [a, b] and f(a) < c < f(b), then ∃x ∈ (a, b): f(x) = c.

Used to prove existence of roots!

2. Extreme Value Theorem

If f is continuous on [a, b], then f attains its maximum and minimum on [a, b].

Fundamental for optimization!

3. Differentiability

If f is differentiable at x₀, then f is continuous at x₀.

Continuity is necessary for differentiability!

4. Integration

Continuous functions on [a, b] are Riemann integrable.

Enables the Fundamental Theorem of Calculus!

Remark 3.10: Big Picture

Continuity is the "niceness" condition that enables most of calculus:

Nice functions have nice properties (IVT, EVT)
Derivatives require continuity
Integration requires "reasonable" behavior
Series convergence theorems often need continuity

15. Quick Reference

Continuity Checklist

To prove f is continuous at x₀:

Verify f(x₀) is defined
Compute lim_{x→x₀} f(x)
Check that limit = f(x₀)

To classify a discontinuity at x₀:

Compute lim_{x→x₀⁺} f(x) and lim_{x→x₀⁻} f(x)
If both finite and equal: Removable (or continuous if = f(x₀))
If both finite but unequal: Jump
If either infinite or DNE: Essential

Important Continuous Functions

Continuous on ℝ

• Polynomials
• sin x, cos x
• eˣ
• |x|

Continuous on Domain

• ln x on (0, ∞)
• tan x where cos x ≠ 0
• √x on [0, ∞)
• arcsin x on [-1, 1]

Discontinuity Classification Table

Type	Left Limit	Right Limit	Example
Removable	L (finite)	L (same)	(x²-1)/(x-1) at x=1
Jump	L₁ (finite)	L₂ ≠ L₁	sgn(x) at x=0
Essential (∞)	±∞	±∞	1/x at x=0
Essential (osc)	DNE	DNE	sin(1/x) at x=0

Operations Preserving Continuity

Operation	Condition	Result
f + g	f, g continuous	Continuous
f - g	f, g continuous	Continuous
f · g	f, g continuous	Continuous
f / g	f, g continuous, g ≠ 0	Continuous
g ∘ f	f cont. at x₀, g cont. at f(x₀)	Continuous
\|f\|	f continuous	Continuous
max(f,g), min(f,g)	f, g continuous	Continuous

What's Next?

In CALC-3.4, we'll explore powerful theorems about continuous functions:

Intermediate Value Theorem: continuous functions take all intermediate values
Extreme Value Theorem: continuous functions on [a,b] attain max and min
Uniform Continuity: same δ works for all points
Heine-Cantor Theorem: continuous on [a,b] ⟹ uniformly continuous

Essential Formulas

Continuity Definition

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

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

Key Identities

\max(f, g) = \frac{f + g + |f - g|}{2}

\min(f, g) = \frac{f + g - |f - g|}{2}

Exam Tips

Common Question Types

• Prove continuity using ε-δ
• Classify discontinuities
• Find constants for continuity
• Analyze piecewise functions
• Prove operations preserve continuity

Key Techniques

• Check three conditions systematically
• Compute one-sided limits first
• Use sequential criterion for non-existence
• Apply triangle inequality for proofs
• Factor to simplify rational functions

Historical Note

The rigorous definition of continuity was developed in the 19th century:

Cauchy (1821): First formal definition using limits
Weierstrass (1860s): The ε-δ definition we use today
Dirichlet (1829): The famous everywhere-discontinuous function
Thomae (1875): The "popcorn function" continuous only at irrationals

These definitions resolved paradoxes about "continuous" curves and made calculus rigorous.

Chapter 3.3 Summary

Continuous at $x_0$ : $\lim f(x) = f(x_0)$
Removable: limit exists but ≠ f(x₀)
Jump: one-sided limits exist but differ
Essential: some one-sided limit doesn't exist
Operations and compositions preserve continuity
Elementary functions are continuous on their domains
Sign preservation: f(x₀) > 0 ⟹ f > 0 nearby
Sequential characterization useful for proving discontinuity

Practice Quiz: Continuity

Questions

Correct

Accuracy

A function

f

is continuous at

x_0

if:

Easy

Not attempted

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

exists but

\neq f(x_0)

, the discontinuity is:

Medium

Not attempted

If both one-sided limits exist but are unequal, the discontinuity is:

Medium

Not attempted

Which function has an essential discontinuity at

x = 0

Hard

Not attempted

f, g

are continuous at

x_0

, which may NOT be continuous at

x_0

Easy

Not attempted

f

is continuous at

x_0

and

g

is continuous at

f(x_0)

, then

g \circ f

is:

Medium

Not attempted

FAQs

What's the intuition behind continuity?

No jumps, no gaps, no holes. You can draw the graph without lifting your pen. Small changes in x produce small changes in f(x).

Why distinguish types of discontinuity?

Removable discontinuities are 'fixable'—just redefine f(x₀). Jump discontinuities appear in step functions. Essential discontinuities show 'wild' behavior.

Are all elementary functions continuous?

Yes, on their natural domains. Polynomials, exponentials, logs, trig functions, and their compositions are continuous where defined.