Types of Discontinuities

7 min read

Reference: Stewart §2.5

Before You Start: Prerequisite Check

📋 Can you do these? (Click to reveal self-test)

Test yourself on these prerequisite skills:

Continuity definition: What are the three conditions for continuity at $a$?

Check

1) $f(a)$ defined, 2) $\lim_{x \to a} f(x)$ exists, 3) They’re equal
One-sided limits: Find $\lim_{x \to 2^-} \frac{\vert x-2\vert }{x-2}$

Check

Answer: $-1$ (when $x < 2$, $\vert x-2\vert = -(x-2)$)
Factoring: Simplify $\frac{x^2 - 9}{x - 3}$ for $x \neq 3$

Check

Answer: $x + 3$ (factor numerator as $(x-3)(x+3)$)

If you struggled:

Review Continuity at a Point before continuing
Review One-Sided Limits for handling different directions

When Continuity Breaks Down

Not all discontinuities are created equal. Some are minor annoyances, like a single hole that could be patched. Others represent fundamental breaks in the function’s behavior, like a staircase step or a vertical asymptote.

Why does this classification matter? Because removable discontinuities can often be fixed by simply redefining the function at one point. Jump discontinuities indicate genuine “breaks” in the function that cannot be repaired. Infinite discontinuities signal unbounded behavior. Knowing which type you’re dealing with tells you what’s possible and what isn’t.

Understanding discontinuity types is also essential for integration (Calculus II) and for understanding where formulas like the Fundamental Theorem of Calculus apply.

Prerequisite Map

Quick Reference

Property	Value
Chapter	1.8
Course	MATH161
Difficulty	Intermediate
Time	~20 minutes

Quick Classification Guide

If you find...	Then it’s a...	Can be fixed?
$\lim$ exists but $\neq f(a)$ or $f(a)$ undefined	Removable	Yes: redefine $f(a) = L$
$\lim_{x \to a^-} \neq \lim_{x \to a^+}$ (both finite)	Jump	No
Either one-sided limit is $\pm\infty$	Infinite	No

Key Concepts

The Three Types

Type	What Happens	Graph Appearance	Can It Be Fixed?
Removable	Limit exists but $\neq f(a)$ or $f(a)$ undefined	Hole in graph	Yes, redefine $f(a)$
Jump	Left and right limits exist but are different	Step or break	No
Infinite	Limit is $\pm\infty$ or doesn’t exist	Vertical asymptote	No

Visual Comparison

REMOVABLE                JUMP                    INFINITE
    │    ╱                   │    ╱                  │    │
    │   ╱                    │   ╱                   │    │ blows up
    │  ○   hole              │  ●────                │    │
    │ ╱                      │       gap             │    │
    │╱                       │────●                  │    │
    ┼────────                ┼────────               ┼────┼────
        a                        a                       a

  Limit exists             Both one-sided         Limit is ±∞
  but ≠ f(a)               limits exist,          or oscillates
                           but differ             without bound

Removable Discontinuity

A discontinuity at $x = a$ is removable if:

$\lim_{x \to a} f(x)$ exists (a finite number $L$)
Either $f(a)$ is undefined, or $f(a) \neq L$

Why “removable”? We can create a new continuous function by defining: $$\tilde{f}(x) = \begin{cases} f(x) & \text{if } x \neq a \\ L & \text{if } x = a \end{cases}$$

Classic example: $f(x) = \dfrac{x^2 - 1}{x - 1}$ at $x = 1$

Factor: $\dfrac{(x-1)(x+1)}{x-1} = x + 1$ for $x \neq 1$

So $\lim_{x \to 1} f(x) = 2$, but $f(1)$ is undefined. Redefine $f(1) = 2$ to fix it.

Jump Discontinuity

A discontinuity at $x = a$ is a jump if:

$\lim_{x \to a^-} f(x) = L_1$ exists (finite)
$\lim_{x \to a^+} f(x) = L_2$ exists (finite)
$L_1 \neq L_2$

The jump height is $\vert L_2 - L_1\vert $.

Cannot be fixed because the function approaches different values from each side.

Classic example: The floor function $f(x) = \lfloor x \rfloor$ at any integer $n$

$\lim_{x \to n^-} \lfloor x \rfloor = n - 1$
$\lim_{x \to n^+} \lfloor x \rfloor = n$
Jump height = 1

Infinite Discontinuity

A discontinuity at $x = a$ is infinite if:

At least one of $\lim_{x \to a^-} f(x)$ or $\lim_{x \to a^+} f(x)$ is $\pm\infty$

This typically occurs at vertical asymptotes.

Classic example: $f(x) = \dfrac{1}{x}$ at $x = 0$

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

Classification Flowchart

graph TD
    A["Is f(a) defined AND<br/>does lim equal f(a)?"] -->|Yes| B["CONTINUOUS"]
    A -->|No| C["Does lim<sub>x→a</sub> f(x) exist<br/>(finite)?"]
    C -->|Yes| D["REMOVABLE<br/>discontinuity"]
    C -->|No| E["Do both one-sided<br/>limits exist (finite)?"]
    E -->|Yes| F["JUMP<br/>discontinuity"]
    E -->|No| G["INFINITE<br/>discontinuity"]

    style B fill:#d1fae5
    style D fill:#fef3c7
    style F fill:#fed7aa
    style G fill:#fecaca

Common Pitfalls

Mistake	Why It’s Wrong	Correct Approach
“All division by zero is infinite”	Canceling factors can make limit finite	Factor first, then check
“Jump means the graph jumps up”	Jump just means left ≠ right limits	Jump can go up OR down
Confusing “hole” with “vertical asymptote”	Hole = removable, asymptote = infinite	Check if limit is finite or infinite
Saying $\frac{1}{x}$ at $x=0$ is “jump”	Both one-sided limits are infinite	Jump requires FINITE one-sided limits

📜 Why Does Classification Matter?

Knowing the type of discontinuity tells you what’s possible:

Removable: Often artifacts of how the function is written. In applications, you’d just “fill the hole.”
Jump: Natural in discrete systems (tax brackets, shipping rates, digital signals).
Infinite: Signal fundamental changes and often mark boundaries of physical validity.

Integration theory (Calculus II) treats these differently: functions with only jump discontinuities can be integrated, but infinite discontinuities require special “improper integral” techniques.

Practice Problems

Level 1 Identifying from Graph Description

A graph has a hole at $(2, 5)$ with no point plotted there, but the function approaches 5 from both sides. What type of discontinuity is this?

Thought Process

A “hole” means the limit exists (both sides approach the same value), but the function is either undefined there or has a different value. This is the hallmark of a removable discontinuity.

Show Answer

Removable discontinuity

The limit $\lim_{x \to 2} f(x) = 5$ exists, but $f(2)$ is undefined (no point plotted). This satisfies the definition of a removable discontinuity.

Level 2 Algebraic Classification

Classify the discontinuity of $f(x) = \dfrac{x^2 - 9}{x - 3}$ at $x = 3$.

Thought Process

The function is undefined at $x = 3$ (division by zero). To classify:

Try to find the limit by factoring the numerator
If the limit exists (finite), it’s removable
If the limit is $\pm\infty$, it’s infinite
If left/right limits differ, it’s a jump

Show Answer

Factor the numerator: $x^2 - 9 = (x-3)(x+3)$

For $x \neq 3$: $$f(x) = \frac{(x-3)(x+3)}{x-3} = x + 3$$

Therefore: $$\lim_{x \to 3} f(x) = \lim_{x \to 3} (x + 3) = 6$$

The limit exists and is finite, but $f(3)$ is undefined.

Classification: Removable discontinuity

(Could be fixed by defining $f(3) = 6$)

Level 2 CCI: Conceptual Classification

A student claims: “If $\lim_{x \to a} f(x) = \infty$, then $f$ has an infinite discontinuity at $a$.”

Is this statement always true, sometimes true, or never true?

(A) Always true

(B) Sometimes true (depends on whether $f(a)$ is defined)

(C) Sometimes true (only if both one-sided limits are infinite)

(D) Never true (infinite limits mean the limit does not exist)

Thought Process

The definition of infinite discontinuity requires at least one of the one-sided limits to be $\pm\infty$.

If $\lim_{x \to a} f(x) = \infty$, this means both one-sided limits equal $+\infty$ (otherwise the two-sided limit wouldn’t exist as $\infty$).

The question is: does this automatically make it an infinite discontinuity?

Show Answer

Answer: (A) Always true

Analysis:

If $\lim_{x \to a} f(x) = \infty$, then at least one one-sided limit is infinite
By definition, this makes $x = a$ an infinite discontinuity
It does not matter whether $f(a)$ is defined. The infinite behavior of the limit determines the type.

Common misconception: Some students think (D) because “the limit doesn’t exist” in the usual finite sense. But we classify based on what the one-sided limits DO, not whether a finite limit exists.

Level 3 Piecewise Discontinuity

Classify all discontinuities of the function: $$g(x) = \begin{cases} x + 2 & \text{if } x < 0 \\ 1 & \text{if } x = 0 \\ x^2 & \text{if } x > 0 \end{cases}$$

Thought Process

Check continuity at the boundary point $x = 0$:

Find $g(0)$
Compute $\lim_{x \to 0^-} g(x)$ using the “$x < 0$” piece
Compute $\lim_{x \to 0^+} g(x)$ using the “$x > 0$” piece
Compare all three values

Show Answer

At $x = 0$:

Step 1: $g(0) = 1$

Step 2: Left-hand limit: $$\lim_{x \to 0^-} g(x) = \lim_{x \to 0^-} (x + 2) = 0 + 2 = 2$$

Step 3: Right-hand limit: $$\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} x^2 = 0$$

Step 4: Since $\lim_{x \to 0^-} g(x) = 2 \neq 0 = \lim_{x \to 0^+} g(x)$, the one-sided limits are different.

Classification: Jump discontinuity at $x = 0$ with jump height $\vert 2 - 0\vert = 2$

Level 4 Multiple Discontinuities

Find and classify all discontinuities of: $$h(x) = \frac{x^2 - 4}{x^2 - 3x + 2}$$

Thought Process

Factor both numerator and denominator to find where the function is undefined
Numerator: $x^2 - 4 = (x-2)(x+2)$
Denominator: $x^2 - 3x + 2 = (x-1)(x-2)$
The function is undefined at $x = 1$ and $x = 2$
Check each: does a common factor cancel (removable) or not (infinite)?

Show Answer

Factor: $$h(x) = \frac{(x-2)(x+2)}{(x-1)(x-2)}$$

The function is undefined at $x = 1$ and $x = 2$.

At $x = 2$:

The factor $(x-2)$ cancels: $$\lim_{x \to 2} h(x) = \lim_{x \to 2} \frac{x+2}{x-1} = \frac{4}{1} = 4$$

The limit exists and is finite → Removable discontinuity at $x = 2$

At $x = 1$:

After canceling: $h(x) = \frac{x+2}{x-1}$ for $x \neq 2$

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

The limits are infinite → Infinite discontinuity at $x = 1$

Level 5 Constructing Functions with Specified Discontinuities

Construct a single function $f(x)$ that has:

A removable discontinuity at $x = -1$
A jump discontinuity at $x = 0$
An infinite discontinuity at $x = 2$

Verify that your function has these properties.

Thought Process

Build the function piecewise or as a rational function:

For removable at $x = -1$: Include a factor $(x+1)$ in both numerator and denominator, or leave a hole

For jump at $x = 0$: Use a piecewise definition with different formulas on each side

For infinite at $x = 2$: Include $(x-2)$ in the denominator but not numerator

One approach: use a piecewise function that incorporates a rational piece.

Show Answer

One solution: $$f(x) = \begin{cases} \dfrac{x+1}{(x+1)(x-2)} & \text{if } x < 0, x \neq -1 \\[8pt] \dfrac{1}{x-2} + 3 & \text{if } x \geq 0 \end{cases}$$

Verification:

At $x = -1$ (removable): For $x < 0$, $x \neq -1$: $f(x) = \dfrac{1}{x-2}$

$\lim_{x \to -1} f(x) = \dfrac{1}{-1-2} = -\dfrac{1}{3}$

The limit exists but $f(-1)$ is undefined → Removable ✓

At $x = 0$ (jump):

$\lim_{x \to 0^-} f(x) = \dfrac{1}{0-2} = -\dfrac{1}{2}$

$\lim_{x \to 0^+} f(x) = \dfrac{1}{0-2} + 3 = -\dfrac{1}{2} + 3 = \dfrac{5}{2}$

Left limit $\neq$ right limit → Jump ✓

At $x = 2$ (infinite):

$\lim_{x \to 2} f(x) = \lim_{x \to 2} \left(\dfrac{1}{x-2} + 3\right) = \pm\infty$

→ Infinite ✓

Mastery Checklist

Novice (Level 1-2):

I can identify the three types of discontinuities from a graph description
I can explain in words what “removable,” “jump,” and “infinite” mean
I can use the classification flowchart correctly

Competent (Level 3):

I can classify discontinuities algebraically by computing one-sided limits
I can calculate the jump height for a jump discontinuity
I can explain why canceling factors can change “infinite” to “removable”

Proficient (Level 4-5):

I can find ALL discontinuities of a rational function and classify each
I can construct functions with specified types of discontinuities
I can explain the implications of each discontinuity type for integration

Mental Model

The “Road Trip” Analogy:

Removable discontinuity: A pothole in the road. Annoying, but you can patch it and drive smoothly.
Jump discontinuity: The road suddenly ends at a cliff, and another road starts at a different height. No amount of paving connects them.
Infinite discontinuity: The road goes vertically up to infinity. There’s no way to drive across.

Connections

Looking back:

Continuity at a point defines what it means for continuity to hold
One-sided limits are essential for detecting jump discontinuities

Looking ahead:

Intermediate Value Theorem applies only to continuous functions
Improper integrals (Calculus II) require understanding infinite discontinuities
Piecewise functions in applications often have jump discontinuities (e.g., tax brackets, shipping rates)

Real-World Example: Toll Road Pricing

🚗 The Toll Road Function

A toll road charges $5 normally but $7 during rush hours (7-10 AM and 4-7 PM). If $T(t)$ is the toll as a function of hours past midnight:

$7  ─────────           ─────────
           │           │
$5  ───────┼───────────┼─────────
           7AM       4PM      7PM

Discontinuity analysis:

At $t = 7$ (7 AM): Jump from $5 to $7 (jump discontinuity)
At $t = 10$ (10 AM): Jump from $7 to $5 (jump discontinuity)
At $t = 16$ (4 PM): Jump from $5 to $7 (jump discontinuity)
At $t = 19$ (7 PM): Jump from $7 to $5 (jump discontinuity)

Significance: If you arrive at 6:59 AM, you pay $5. At 7:01 AM, you pay $7. The discontinuity represents an abrupt policy change with no smooth transition.

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Continuity at a Point	Skills Index	Continuity of Combined Functions

Last updated: 2026-01-22