Exploring types of discontinuities

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

1. Continuity and Discontinuity Basics★★☆☆☆⏱ 3 min

A function is discontinuous at $x = a$ if it fails any of the three conditions for continuity at $a$ . Exploring types of discontinuities is the process of classifying discontinuities based on which condition fails, and the behavior of the function’s limits at that point. This topic makes up 10–12% of the AP Calculus AB exam weight, and is a prerequisite for topics like the Intermediate Value Theorem, differentiability, and integration.

📘 Definition

Continuity at a Point

A function $f (x)$ is continuous at $x = a$ if and only if all three conditions are satisfied: $f (a)$ is defined, $l i m_{x \to a} f (x)$ exists, and $l i m_{x \to a} f (x) = f (a)$ . If any condition fails, the function is discontinuous at $x = a$ .

2. Removable Discontinuities★★☆☆☆⏱ 4 min

📘 Definition

Removable Discontinuity

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

Removable discontinuities most commonly occur in rational functions when a common linear factor cancels from the numerator and denominator, leaving a "hole" in the graph. They can also occur in piecewise functions when the defined function value at a breakpoint does not match the approaching limit.

📐 Worked Example

Classify the discontinuity of $f (x) = \frac{x ^{2} - 16}{x - 4}$ at $x = 4$ .

1
Check the first continuity condition: the denominator is 0 when $x = 4$ , so $f (4)$ is undefined, meaning $f (x)$ is discontinuous at $x = 4$ .
2
Simplify the function for $x \neq = 4$ by factoring the numerator:
3
$\frac{x^2 - 16}{x - 4} = \frac{(x-4)(x+4)}{x-4} = x+4 \quad \text{for } x\neq 4$
4
Evaluate the two-sided limit:
5
$\lim_{x\to 4} (x+4) = 8$
6
The limit is finite and exists. Since the two-sided limit exists but the function is undefined at $x = 4$ , this is a removable discontinuity. Redefining $f (4) = 8$ makes the function continuous at $x = 4$ .

3. Jump Discontinuities★★★☆☆⏱ 3 min

📘 Definition

Jump Discontinuity

A non-removable discontinuity at $x = a$ where both left-hand and right-hand limits exist as finite values, but are not equal. The two-sided limit $l i m_{x \to a} f (x)$ does not exist, so the discontinuity cannot be fixed by redefining $f (a)$ .

Jump discontinuities are most common in piecewise functions with different expressions on either side of a breakpoint, and in step functions like the floor or ceiling functions. The value of $f (a)$ (whether defined or not) does not change the classification, as long as the one-sided limits are finite and unequal.

📐 Worked Example

Classify the discontinuity of $f (x) = {3 x - 2 x^{2} + 1 x < 1 x > 1$ at $x = 1$ .

1
First, confirm $f (x)$ is discontinuous: $f (1)$ is not defined by the piecewise rule, so it is discontinuous at $x = 1$ .
2
Evaluate the left-hand limit using the left-side expression:
3
$\lim_{x\to 1^-} (3x - 2) = 3(1) - 2 = 1$
4
Evaluate the right-hand limit using the right-side expression:
5
$\lim_{x\to 1^+} (x^2 + 1) = (1)^2 + 1 = 2$
6
Both one-sided limits are finite, but $1 \neq = 2$ , so they are not equal. This is a jump discontinuity. Even if we defined $f (1)$ as 1 or 2, the two-sided limit would still not exist, so it remains non-removable.

4. Infinite Discontinuities★★★☆☆⏱ 4 min

📘 Definition

Infinite Discontinuity

A non-removable discontinuity at $x = a$ where at least one one-sided limit approaches $\

For rational functions, infinite discontinuities occur when the denominator is zero at $x = a$ , but the numerator is non-zero at $x = a$ , so the fraction grows without bound as $x$ approaches $a$ . If the factor causing the zero denominator does not cancel with any factor in the numerator, the discontinuity is infinite.

📐 Worked Example

Classify the discontinuity of $f (x) = \frac{x + 4}{x ^{2} - 3 x + 2}$ at $x = 2$ .

1
Factor the denominator: $x^{2} - 3 x + 2 = (x - 1) (x - 2)$ , so $f (2)$ has a zero denominator and is undefined. The numerator at $x = 2$ is $2 + 4 = 6 \neq = 0$ .
2
Evaluate the left-hand limit as $x \to 2^{-}$ : $x - 2$ is negative, $(x - 1)$ is positive, so the fraction approaches:
3
$-\infty$
4
Evaluate the right-hand limit as $x \to 2^{+}$ : $x - 2$ is positive, so the fraction approaches:
5
$+\infty$
6
Since at least one one-sided limit is infinite, this is an infinite discontinuity.

✓ Quick check

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

Let $f (x) = ⎩ ⎨ ⎧ \frac{x ^{2} - 4 x + 3}{x - 3} 0 x \neq = 3 x = 3$ . Which correctly classifies the discontinuity at $x = 3$ ?
- Infinite discontinuity
- Jump discontinuity
- Removable discontinuity
- $f$ is continuous at $x = 3$
Reveal answer
Removable discontinuity —
Factor the numerator to get $(x - 3) (x - 1)$ , cancel to get $l i m_{x \to 3} f (x) = 2$ . $f (3) = 0 \neq = 2$ , so the two-sided limit exists and this is a removable discontinuity.

5. Common Pitfalls

Wrong move:

Classifying any undefined point of a rational function as a removable discontinuity

Why:

Students assume all undefined points are holes, forgetting that only discontinuities with a finite existing limit are removable.

Correct move:

Always evaluate the limit at the undefined point before classifying; if the limit is infinite, it is an infinite discontinuity.

Wrong move:

Calling a jump discontinuity removable just because $f (a)$ is undefined

Why:

Students confuse the requirement of an existing finite two-sided limit for removable discontinuities with just the function being undefined.

Correct move:

Check if left and right limits are equal first; if they are not equal, it is a jump discontinuity regardless of whether $f (a)$ is defined.

Wrong move:

Concluding $f (x)$ is continuous at $x = a$ just because $f (a)$ is defined

Why:

Students forget the other two continuity conditions: the limit must exist, and the limit must equal the function value.

Correct move:

Always check all three continuity conditions in order before concluding if the function is continuous or what type of discontinuity it has.

Wrong move:

Stating the two-sided limit exists for a jump discontinuity because both one-sided limits are finite

Why:

Students confuse 'one-sided limits exist' with 'two-sided limit exists'.

Correct move:

Remember the two-sided limit exists only if both one-sided limits exist and are equal, so jump discontinuities never have a two-sided limit.

Wrong move:

Classifying a discontinuity as infinite just because the function is undefined at that point

Why:

Students do not factor rational functions completely, so they miss common factors that cancel to create a removable discontinuity.

Correct move:

Always factor numerator and denominator completely, and cancel common factors before classifying.

6. Quick Reference Cheatsheet

Category	Definition/Rule	Notes
Continuous at $x = a$	$f (a)$ defined, $l i m_{x \to a} f (x)$ exists, $l i m_{x \to a} f (x) = f (a)$	All three conditions must be satisfied
Removable Discontinuity	$lim_{x \to a} f (x)$ exists finite, but $f (a)$ undefined or $\
Jump Discontinuity	$\
Infinite Discontinuity	At least one one-sided limit is $\
Rational function classification	If $(x - a)$ cancels from numerator/denominator: removable. If not: infinite	Always factor completely first
Discontinuity at piecewise breakpoints	Always evaluate left limit from the left expression, right limit from the right expression	Never assume both sides share the same limit
$f (a)$ value does not change classification	Type of discontinuity depends only on limits, not the value of $f (a)$	Even if $f (a)$ is defined, classification is based on limits

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.

2023 · MCQ
Classify discontinuity of rational function
2022 · FRQ
Identify and classify discontinuities

What's Next

Mastering discontinuity classification is a foundational prerequisite for all work on continuity and its applications across AP Calculus AB. Immediately after this topic, you will apply your understanding to the Intermediate Value Theorem, which requires confirming continuity on an interval before you can apply the theorem’s conclusion. You will also use this classification when checking differentiability, since a function can never be differentiable at a point of discontinuity, so identifying discontinuities is the first step in checking differentiability. Later, when integrating piecewise or rational functions, you will need discontinuity classification to set up integrals correctly and solve net area problems.