Exploring types of discontinuities

AP Calculus BC· AP Calculus BC CED — Limits and Continuity· 14 min read

1. Overview of Discontinuity Classification★☆☆☆☆⏱ 2 min

Discontinuities occur when any of the three conditions for continuity at $x = a$ are violated: (1) $f (a)$ is defined, (2) $lim_{x \to a} f (x)$ exists, (3) $lim_{x \to a} f (x) = f (a)$ . This topic requires you to systematically categorize the type of discontinuity, not just identify that a discontinuity exists.

Discontinuities are first split into removable (can be fixed by redefining one point) and non-removable (cannot be fixed by redefining one point). Non-removable discontinuities are further split into jump, infinite, and oscillating types. This topic makes up 4-6% of the AP Calculus BC exam, appearing in both multiple-choice and free-response questions.

📘 Definition

Discontinuity at a point

A violation of one or more of the three conditions for continuity at $x = a$ , resulting in a break in the graph of $f (x)$ at $a$ .

Example:

All points where a rational function has a zero denominator are candidate discontinuities.

2. Removable Discontinuities★★☆☆☆⏱ 3 min

📘 Definition

Removable Discontinuity

Also called a hole

A discontinuity at $x = a$ where the two-sided limit $lim_{x \to a} f (x)$ exists and is finite, but either $f (a)$ is undefined, or $f (a)$ is defined and not equal to the limit. The discontinuity can be removed by redefining $f (a)$ to equal the limit.

Example:

Occurs at cancelled common factors in rational functions

Removable discontinuities most commonly occur in rational functions when a common factor cancels out from the numerator and denominator. They can also occur in piecewise functions where the defined point does not match the surrounding limit.

📐 Worked Example

Classify the discontinuity of $f (x) = \frac{x ^{2} - 9}{x ^{2} - 2 x - 3}$ at $x = 3$ , if a discontinuity exists.

1
Check the first continuity condition at $a = 3$ : The denominator at $x = 3$ is $3^{2} - 2 (3) - 3 = 0$ , so $f (3)$ is undefined, meaning a discontinuity exists.
2
Factor the numerator and denominator to evaluate the limit:
3
$x^2 -9 = (x-3)(x+3), \quad x^2 - 2x -3 = (x-3)(x+1)$
4
The function simplifies to $\frac{( x - 3 ) ( x + 3 )}{( x - 3 ) ( x + 1 )}$ for all $x \neq = 3$ . Cancel the common factor $(x - 3)$ , which is valid because $x \neq = 3$ when taking the limit, to get:
5
$\lim_{x \to 3} f(x) = \lim_{x \to 3} \frac{x+3}{x+1} = \frac{6}{4} = \frac{3}{2}$
6
The two-sided limit exists and is finite, but $f (3)$ is undefined, so this is a removable discontinuity.

3. Non-Removable Discontinuities: Jump★★☆☆☆⏱ 3 min

📘 Definition

Jump Discontinuity

A non-removable discontinuity at $x = a$ where both one-sided limits exist and are finite, but are not equal to each other. The two-sided limit does not exist, so you cannot redefine $f (a)$ to make the function continuous.

Example:

Occurs in piecewise functions, absolute value functions, and step functions like $⌊ x ⌋$

Jump discontinuities get their name from the 'jump' the function makes from one value to another at the point $a$ . Because both one-sided limits are finite but unequal, there is no way to choose a single value for $f (a)$ that will make the two-sided limit exist, hence the discontinuity is non-removable.

📐 Worked Example

Given $g (x) = \frac{∣ x ^{2} - 4∣}{x - 2}$ , classify the discontinuity at $x = 2$ .

1
Rewrite the absolute value as a piecewise function. First, factor $x^{2} - 4 = (x - 2) (x + 2)$ . For $x > 2$ , $x^{2} - 4 > 0$ , so $∣ x^{2} - 4∣ = x^{2} - 4$ ; for $x < 2$ , $x^{2} - 4 < 0$ , so $∣ x^{2} - 4∣ = - (x^{2} - 4)$ .
2
Evaluate the left-hand limit:
3
$\lim_{x \to 2^-} \frac{-(x-2)(x+2)}{x-2} = \lim_{x \to 2^-} -(x+2) = -4$
4
Evaluate the right-hand limit:
5
$\lim_{x \to 2^+} \frac{(x-2)(x+2)}{x-2} = \lim_{x \to 2^+} (x+2) = 4$
6
Both one-sided limits exist and are finite, but they are not equal, so the two-sided limit does not exist. This matches the definition of a jump discontinuity, a non-removable discontinuity.

4. Non-Removable Discontinuities: Infinite★★☆☆☆⏱ 3 min

📘 Definition

Infinite Discontinuity

A non-removable discontinuity at $x = a$ where at least one of the one-sided limits as $x \to a$ is infinite ( $+ \infty$ or $- \infty$ ).

Example:

Corresponds to vertical asymptotes, where an uncancelled factor makes the denominator zero in a rational function

Infinite discontinuities occur when a factor of the denominator does not cancel with a matching factor in the numerator of a rational function. Because the limit is not finite, the discontinuity cannot be removed by redefining $f (a)$ .

📐 Worked Example

Classify all discontinuities of $h (x) = \frac{x + 5}{x ^{2} - 25}$ .

1
Factor the denominator to find all points where $h (x)$ is undefined: $x^{2} - 25 = (x - 5) (x + 5)$ , so $h (x)$ is undefined at $x = 5$ and $x = - 5$ , so both are candidate discontinuities.
2
Evaluate the limit at $x = - 5$ first:
3
$\lim_{x \to -5} \frac{x+5}{(x-5)(x+5)} = \lim_{x \to -5} \frac{1}{x-5} = -\frac{1}{10}$
4
The limit exists and is finite, but $h (- 5)$ is undefined, so $x = - 5$ is a removable discontinuity.
5
Evaluate the one-sided limits at $x = 5$ :
6
$\lim_{x \to 5^-} \frac{1}{x-5} = -\infty \quad \text{and} \quad \lim_{x \to 5^+} \frac{1}{x-5} = +\infty$
7
At least one one-sided limit is infinite, so $x = 5$ is an infinite non-removable discontinuity.

5. Non-Removable Discontinuities: Oscillating★★★☆☆⏱ 2 min

📘 Definition

Oscillating Discontinuity

A non-removable discontinuity at $x = a$ where the function oscillates infinitely many times as $x \to a$ , never settling to a single finite value, and does not approach $\pm \infty$ , so the two-sided limit does not exist.

Example:

Almost always occurs for trigonometric functions of $\frac{1}{x - a}$ , like $sin (\frac{1}{x})$ at $x = 0$

This is the least frequently tested type of discontinuity on the AP exam, but it is still examinable, most often in multiple-choice questions asking to identify the type from a description or graph.

📐 Worked Example

The function $f (x) = cos (\frac{2}{x + 1})$ has a discontinuity at $x = - 1$ . What type of discontinuity is it?

1
Check $f (- 1)$ : it is undefined, since the argument of cosine is $\frac{2}{0}$ , which is undefined, so a discontinuity exists.
2
Analyze behavior as $x \to - 1$ : as $x \to - 1$ , $\frac{2}{x + 1} \to \pm \infty$ , so $cos (\frac{2}{x + 1})$ oscillates between $- 1$ and $1$ infinitely many times, never approaching a single finite limit, and does not approach $\pm \infty$ .
3
The two-sided limit does not exist, it is not jump (one-sided limits are not finite and unequal) and not infinite (the function does not diverge to infinity). This is an oscillating non-removable discontinuity.

✓ Quick check

Test your understanding with this AP-style multiple-choice question:

How many of the following have a non-removable discontinuity at the given point?
I. $f (x) = \frac{x ^{2} - 4 x}{x ^{2} - 16}$ at $x = 4$
II. $g (x) = \frac{⌊ x ^{2} ⌋}{x - 2}$ at $x = 2$
III. $h (x) = (x - 3) e^{1/ (x - 3)}$ at $x = 3$
- (A) 0
- (B) 1
- (C) 2
- (D) 3
Reveal answer
(C) —
I has a removable discontinuity, II and III are non-removable, so 2 of 3 are non-removable.

6. Common Pitfalls

Wrong move:

Classify $f (x) = \frac{( x - 2 ) ( x + 1 )}{x - 2}$ as having an infinite discontinuity at $x = 2$ , because the denominator is zero.

Why:

Students associate zero denominator with vertical asymptotes without factoring and checking for common factors.

Correct move:

Always factor numerator and denominator completely, cancel common factors, and evaluate the limit at the point before classifying.

Wrong move:

Classify a jump discontinuity at $x = a$ as removable because $f (a)$ is undefined.

Why:

Students confuse the existence of a defined $f (a)$ with the requirement that the two-sided limit must exist for a discontinuity to be removable.

Correct move:

First check if the two-sided limit exists; even if $f (a)$ is undefined, if the two-sided limit doesn't exist, the discontinuity is non-removable.

Wrong move:

Claim $lim_{x \to 0} sin (1/ x) = 0$ , so it is a removable discontinuity, because the function oscillates around zero.

Why:

Students mistake the midpoint of oscillation for the limit.

Correct move:

If the function oscillates infinitely many times near $x = a$ without settling to a single value, the limit does not exist, and the discontinuity is oscillating (non-removable).

Wrong move:

For $f (x) = ⎩ ⎨ ⎧ x^{2} 1 3 x - 2 x < 1 x = 1 x > 1$ , claim $x = 1$ has a jump discontinuity because $f (1) = 1$ doesn't match the expected value.

Why:

Students forget to check that one-sided limits are equal before classifying.

Correct move:

Always evaluate left and right limits first; if both are equal, the discontinuity is removable regardless of the value of $f (a)$ .

Wrong move:

State that any discontinuity where $f (a)$ is undefined is removable.

Why:

Students overgeneralize the case of cancelled common factors.

Correct move:

Only classify as removable if the two-sided limit at $a$ is finite, regardless of whether $f (a)$ is defined or not.

7. Quick Reference Cheatsheet

Discontinuity Type	Conditions at $x = a$	Key Notes
Removable	$lim_{x \to a} f (x)$ exists/finite; $f (a)$ undefined OR $f (a) \neq = lim$	Removable by redefining $f (a) = lim$ ; occurs at cancelled common factors
Jump (Non-removable)	Both one-sided limits exist/finite, but not equal	Occurs in piecewise, absolute value, step functions
Infinite (Non-removable)	At least one one-sided limit is $\pm \infty$	Corresponds to vertical asymptotes; uncancelled denominator zero
Oscillating (Non-removable)	Infinite oscillation near $a$ , limit not finite or infinite	Almost always trig $(1/ (x - a))$ ; rarely tested
Continuity	$f (a)$ defined; 2. $lim_{x \to a} f (x)$ exists; 3. $lim = f (a)$	Discontinuity if any condition fails

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · MCQ
Classify discontinuity types
2019 · FRQ
Justify discontinuity classification

Going deeper

official syllabusAP Calculus BC CEDOfficial College Board course guide

What's Next

Classifying discontinuities is a foundational skill for AP Calculus BC that supports nearly all future units across the course. Understanding discontinuity behavior is required when applying the Intermediate Value Theorem, finding vertical asymptotes for curve sketching, evaluating improper integrals, and analyzing points of differentiability. Mastery of this topic helps you avoid common errors in justifying conclusions about continuity and limits on FRQ questions, where proper justification is worth a significant portion of the points. Next, you will build on this knowledge to explore additional core continuity concepts and their real-world and theoretical applications.