United StatesCalculusSyllabus dot point

What are the different ways a function can fail to be continuous, and how do you tell them apart?

Topic 1.10 Exploring Types of Discontinuities: classify discontinuities as removable (point/hole), jump, or infinite (asymptotic), using limits.

A focused answer to AP Calculus AB Topic 1.10, classifying removable (hole), jump and infinite (asymptotic) discontinuities using one-sided and two-sided limits, with worked identification.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this topic is asking
The three types
How to diagnose
Reading them on a graph
Connecting the type to the continuity conditions
Why the distinction matters

What this topic is asking

The College Board (Topic 1.10) wants you to classify the ways a function can break, using limits. There are three standard types: removable (a hole or point discontinuity), jump, and infinite (asymptotic). Each is diagnosed by comparing the one-sided limits, the two-sided limit, and the function value.

The three types

How to diagnose

This procedure works straight from a formula, a graph, or a table.

Reading them on a graph

A removable discontinuity looks like an open circle (a hole), sometimes with a stray filled dot elsewhere. A jump looks like two pieces at different heights with a gap. An infinite discontinuity looks like the curve shooting up or down along a vertical line. Recognizing the picture is half the battle on multiple-choice questions.

Classifying a discontinuity from a formula

Classify the discontinuity of $f(x) = \frac{x - 1}{x^2 - 1}$ at $x = 1$ and at $x = -1$ .

step 1 Factor

$x^2 - 1 = (x-1)(x+1)$ , so $f(x) = \frac{x-1}{(x-1)(x+1)} = \frac{1}{x+1}$ for $x \neq 1$ .

step 2 Examine $x = 1$

The factor $(x - 1)$ cancels, so $\lim_{x \to 1} f(x) = \frac{1}{1+1} = \frac{1}{2}$ , which is finite, but $f(1)$ is undefined. This is a removable discontinuity (a hole at $\left(1, \tfrac12\right)$ ).

step 3 Examine $x = -1$

The factor $(x + 1)$ does not cancel. As $x \to -1$ , the denominator $\to 0$ while the numerator $\to -2 \neq 0$ , so $\lim_{x \to -1} f(x) = \pm\infty$ . This is an infinite discontinuity (a vertical asymptote at $x = -1$ ).

step 4 Summarize

$f$ has a removable discontinuity at $x = 1$ and an infinite discontinuity at $x = -1$ .

Connecting the type to the continuity conditions

Each discontinuity type corresponds to a specific way the three-part continuity test fails, which is a clean way to keep them straight. A removable discontinuity satisfies "the limit exists" but fails "the value equals the limit" (the value is missing or sits at the wrong height) - only the third condition breaks. A jump fails "the limit exists" because the one-sided limits disagree, so the second condition breaks. An infinite discontinuity also fails "the limit exists", but because a one-sided limit is unbounded rather than merely mismatched. Reading a discontinuity through the lens of which continuity condition it violates both classifies it and explains why it is discontinuous, which is exactly the reasoning a free-response question wants.

Why the distinction matters

The type tells you what you can do next. A removable discontinuity can be "fixed" by redefining the single point (Topic 1.13). A jump or infinite discontinuity cannot be removed - the function genuinely fails to settle on one value. This classification also feeds straight into continuity at a point (Topic 1.11) and the Intermediate Value Theorem (Topic 1.16), which require continuity. For a rational function, the practical workflow is to factor numerator and denominator, cancel any shared factors (each cancelled factor marks a removable hole), and treat the surviving denominator zeros as infinite discontinuities; jumps, by contrast, arise from piecewise definitions rather than from rational expressions. This factor-first routine classifies every discontinuity of a rational function in one pass.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2021 (style)1 marksSection I (multiple choice). The function

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

has what kind of discontinuity at

x = 2

? (A) Jump (B) Infinite (C) Removable (D) None

Show worked answer →

The correct answer is (C), removable.

For $x \neq 2$ , $f(x) = \frac{(x-2)(x+2)}{x-2} = x + 2$ , so $\lim_{x \to 2} f(x) = 4$ exists (it is finite), but $f(2)$ is undefined. A discontinuity where the two-sided limit exists but the function value is missing or different is a removable discontinuity (a hole). It is not a jump (the one-sided limits agree) and not infinite (the limit is finite).

AP 2023 (style)3 marksSection II (free response). Classify the discontinuity of each function at the indicated point and justify with limits: (a)

f(x) = \frac{1}{x - 3}

x = 3

. (b) the step function with

g(x) = 1

for

x < 0

and

g(x) = 2

for

x \geq 0

, at

x = 0

. (c)

h(x) = \frac{x^2 - 1}{x - 1}

x = 1

Show worked answer →

A 3-point classification question.

(a) (1 point) Infinite discontinuity: $\lim_{x \to 3^+} \frac{1}{x-3} = +\infty$ and $\lim_{x \to 3^-} = -\infty$ , so the limit is infinite (a vertical asymptote).
(b) (1 point) Jump discontinuity: $\lim_{x \to 0^-} g(x) = 1$ and $\lim_{x \to 0^+} g(x) = 2$ ; the one-sided limits are finite but unequal.
(c) (1 point) Removable discontinuity: $h(x) = x + 1$ for $x \neq 1$ , so $\lim_{x \to 1} h(x) = 2$ exists but $h(1)$ is undefined (a hole).

Related dot points

Sources & how we know this

AP Calculus AB and BC Course and Exam Description — College Board (2020)