Exploring types of discontinuities — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: Classifying removable, jump, and infinite discontinuities, applying the three continuity conditions, identifying discontinuities in rational, piecewise, and contextual functions, and redefining removable discontinuities for continuity.

You should already know: How to evaluate one-sided and two-sided limits. How to factor and simplify rational functions. How to evaluate piecewise function values at breakpoints.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus AB style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Exploring types of discontinuities?

A function is discontinuous at $x=a$ if it fails any of the three conditions for continuity at $a$: 1) $f(a)$ is defined, 2) ${\lim }_{x\to a}f(x)$ exists, and 3) ${\lim }_{x\to a}f(x)=f(a)$. Exploring types of discontinuities is the process of classifying the discontinuity based on which condition fails, and what the behavior of the function’s limits is at that point. This topic is part of Unit 1: Limits and Continuity, which makes up 10–12% of the AP Calculus AB exam weight. It appears in both multiple-choice (MCQ) and free-response (FRQ) sections, both as standalone classification questions and as a prerequisite for questions about continuity, the Intermediate Value Theorem, and differentiability. Mastery of discontinuity classification is required for almost every subsequent unit of AP Calculus AB, as derivatives and integrals behave differently at discontinuous points.

2. Removable Discontinuities (Point Discontinuities)

A removable discontinuity at $x=a$ is defined by the existence of a finite two-sided limit ${\lim }_{x\to a}f(x)$, but a failure of either the first or third continuity condition: either $f(a)$ is undefined, or $f(a)$ is defined but does not equal the limit. The name "removable" comes from the fact that you can remove the discontinuity by redefining $f(a)$ to equal the limit, which makes the function continuous at $a$.

The most common location of removable discontinuities is rational functions, where a linear factor cancels out from the numerator and denominator, leaving a "hole" in the graph at the root of the canceled factor. Removable discontinuities can also occur in piecewise functions where the defined value of the function at the 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. First, 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 $x\ne 4$ by factoring the numerator: $$\frac{x^2-16}{x-4}=\frac{(x-4)(x+4)}{x-4}=x+4\text{for }x\ne 4$$
3. Evaluate the two-sided limit: ${\lim }_{x\to 4}(x+4)=8$, which is finite and exists.
4. Since the two-sided limit exists but the function is undefined at $x=4$, this is a removable discontinuity. If we redefine $f(4)=8$, the function becomes continuous at $x=4$.

Exam tip: Always simplify rational functions completely before classifying discontinuities — any factor that cancels from numerator and denominator always produces a removable discontinuity at the corresponding root.

3. Jump Discontinuities

A jump discontinuity at $x=a$ is a non-removable discontinuity where both the left-hand and right-hand limits exist as finite values, but the two one-sided limits are not equal. Because the left and right limits do not match, the two-sided limit ${\lim }_{x\to a}f(x)$ does not exist, so the discontinuity cannot be fixed by redefining $f(a)$. Visually, the graph of the function "jumps" from one finite value to another at $x=a$, which is where the name comes from.

Jump discontinuities are most common in piecewise functions that have different expressions on either side of a breakpoint, and in step functions like the greatest integer (floor) function. The value of $f(a)$ itself (whether it is defined, and what it equals) does not affect the classification: even if $f(a)$ is defined, as long as the one-sided limits are finite and unequal, the discontinuity is still a jump discontinuity.

Worked Example

Classify the discontinuity of $f(x)=\{\begin{array}{ll}3x-2& x<1\\ x^2+1& x>1\end{array}$ 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: ${\lim }_{x\to 1^{-}}(3x-2)=3(1)-2=1$.
3. Evaluate the right-hand limit: ${\lim }_{x\to 1^{+}}(x^2+1)=(1{)}^2+1=2$.
4. Both one-sided limits are finite, but $1\ne 2$, so they are not equal. This fits the definition of a jump discontinuity. Even if we defined $f(1)$ as 1 or 2, the two-sided limit would still not exist, so it remains a non-removable jump discontinuity.

Exam tip: When classifying discontinuities at piecewise function breakpoints, always evaluate the left limit using the left-side expression and the right limit using the right-side expression — never plug the breakpoint into only one side of the function.

4. Infinite Discontinuities

An infinite discontinuity at $x=a$ is a non-removable discontinuity where at least one of the one-sided limits approaches $\pm \infty$. Unlike removable and jump discontinuities, the function does not approach a finite value from at least one side, so no redefinition of $f(a)$ can make the function continuous here. Infinite discontinuities occur exactly at vertical asymptotes of a function.

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

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

Exam tip: A common trick on AP exams is to write a rational function with multiple discontinuities — always check each zero of the denominator separately to avoid misclassifying one as removable when it is actually infinite.

5. Common Pitfalls (and how to avoid them)

• 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. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

Let $f(x)=\{\begin{array}{ll}\frac{x^2-4x+3}{x-3}& x\ne 3\\ 0& x=3\end{array}$. Which of the following correctly classifies the discontinuity at $x=3$? A) Infinite discontinuity B) Jump discontinuity C) Removable discontinuity D) $f$ is continuous at $x=3$

Worked Solution: First, factor the numerator of the expression for $x\ne 3$: $x^2-4x+3=(x-3)(x-1)$. Cancel the $(x-3)$ term to get $f(x)=x-1$ for $x\ne 3$. Evaluate the two-sided limit: ${\lim }_{x\to 3}(x-1)=2$. $f(3)$ is defined as 0, so the limit $2\ne f(3)=0$. The two-sided limit exists and is finite, so this is a removable discontinuity. We eliminate A (limit is finite, not infinite), B (left and right limits are equal), D (the limit does not equal the function value, so it is discontinuous). The correct answer is C.

Question 2 (Free Response)

Let $g(x)=\frac{3x^2-12}{x^2-x-2}$. (a) Find all values of $x$ where $g(x)$ is discontinuous. (b) Classify each discontinuity from (a) as removable, jump, or infinite. (c) What value of $g(a)$ must be assigned to the $x$-coordinate of the removable discontinuity to make $g(x)$ continuous at that point?

Worked Solution: (a) First, factor the denominator: $x^2-x-2=(x-2)(x+1)$. $g(x)$ is undefined when the denominator is zero, so the discontinuities are at $x=2$ and $x=-1$. (b) Factor the numerator: $3x^2-12=3(x^2-4)=3(x-2)(x+2)$. Rewrite $g(x)$ for $x\ne 2$: $$g(x)=\frac{3(x-2)(x+2)}{(x-2)(x+1)}=\frac{3(x+2)}{x+1}$$ At $x=2$, the limit ${\lim }_{x\to 2}g(x)=\frac{3(4)}{3}=4$, which is finite, so $x=2$ is a removable discontinuity. At $x=-1$, the $(x+1)$ factor does not cancel, the numerator at $x=-1$ is $3(1)=3\ne 0$, so one or more one-sided limits are infinite. Thus $x=-1$ is an infinite discontinuity. There are no jump discontinuities. (c) To make $g(x)$ continuous at $x=2$, we set $g(2)={\lim }_{x\to 2}g(x)=4$, so the required value is $g(2)=4$.

Question 3 (Application / Real-World Style)

A small coffee shop charges $4.50forthefirst12ouncesofcoldbrew,plus$0.30 for each additional ounce (rounded up to the next full ounce). The total cost $C(w)$ for $w$ ounces of cold brew is $C(w)=4.50+0.30\lceil w-12\rceil$ for $w>0$, where $\lceil z\rceil$ is the ceiling function (the smallest integer greater than or equal to $z$). Identify all discontinuities of $C(w)$ for $w>12$, classify each type, and explain the context.

Worked Solution: The ceiling function $\lceil w-12\rceil$ changes value at every integer $w=12+n$, where $n=1,2,3,...$ (i.e., $w=13,14,15,...$). Evaluate the one-sided limits at any $w=12+n$: ${\lim }_{w\to (12+n{)}^{-}}\lceil w-12\rceil =n-1$, so ${\lim }_{w\to (12+n{)}^{-}}C(w)=4.50+0.30(n-1)=4.20+0.30n$. ${\lim }_{w\to (12+n{)}^{+}}\lceil w-12\rceil =n$, so ${\lim }_{w\to (12+n{)}^{+}}C(w)=4.50+0.30n$. Both one-sided limits are finite, but they differ by $0.30$, so $C(w)$ has a jump discontinuity at every integer $w>12$. In context, this means the total cost jumps by $0.30 when you cross into a new whole ounce of cold brew, matching the coffee shop's pricing rule.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundational prerequisite for all work on continuity and its applications in AP Calculus AB. Immediately after classifying discontinuities, 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 handle improper integrals (for BC candidates, though AB candidates still need this for net area problems).

• Intermediate Value Theorem
• Continuity over intervals
• Differentiability and continuity
• Integration of piecewise functions