Exploring types of discontinuities — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Removable discontinuities (holes), jump discontinuities, infinite discontinuities, and oscillating discontinuities; classification of discontinuities from graphs and algebraic expressions, using continuity definitions to categorize discontinuity type per AP CED standards.

You should already know: How to evaluate one-sided and two-sided limits at a point, the formal definition of continuity at a point, basic algebraic simplification for rational and piecewise functions.

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 BC 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?

This topic, required by AP Calculus CED Unit 1 (Limits and Continuity), makes up roughly 4-6% of the AP Calculus BC exam weight. It centers on systematically categorizing breaks in the continuity property of a function at a given point, rather than just identifying that a discontinuity exists. Discontinuities occur when any of the three conditions for continuity at $x=a$ are violated: $f(a)$ is defined, ${\lim }_{x\to a}f(x)$ exists, and ${\lim }_{x\to a}f(x)=f(a)$. When one or more conditions fail, we sort the discontinuity by which condition fails and the behavior of the one-sided limits at the point. This topic appears in both multiple-choice (MCQ) questions, where you may count or classify discontinuities of a given function, and free-response questions (FRQ), where you may need to justify discontinuity classification as part of larger problems involving derivatives, integrals, or the Intermediate Value Theorem. It is a foundational topic that builds directly on limit evaluation and sets up for later concepts across the course.

2. Removable Discontinuities

A removable discontinuity (also called a hole) occurs at $x=a$ when 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 but not equal to the limit. The name "removable" comes from the fact that we can redefine $f(a)$ to equal the existing limit, which "fills in" the hole and makes the function continuous at $a$. This most commonly occurs in rational functions where a common factor cancels out from the numerator and denominator. For example, $f(x)=\frac{(x-2)(x+3)}{x-2}$ has a common factor of $(x-2)$, so the function is undefined at $x=2$, but the limit as $x\to 2$ is $5$, so we can redefine $f(2)=5$ to remove the discontinuity. Another common case is a piecewise function where the point is defined incorrectly relative to the surrounding trend.

Worked Example

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

1. First 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: $x^2-9=(x-3)(x+3)$, $x^2-2x-3=(x-3)(x+1)$, so $f(x)=\frac{(x-3)(x+3)}{(x-3)(x+1)}$ for all $x\ne 3$.
3. Cancel the common factor $(x-3)$ (valid for $x\to 3$, since $x\ne 3$ when taking the limit) to get: $$\underset{x\to 3}{\lim }f(x)=\underset{x\to 3}{\lim }\frac{x+3}{x+1}=\frac{6}{4}=\frac{3}{2}$$
4. The two-sided limit exists and is finite, but $f(3)$ is undefined, so this is a removable discontinuity.

Exam tip: If you cancel a common factor from a rational function, any $x$-value that made that factor zero is always a removable discontinuity, not another type. Always confirm the limit is finite before classifying it as removable.

3. Non-Removable Discontinuities: Jump Discontinuities

A non-removable discontinuity is any discontinuity where the two-sided limit does not exist, so you cannot redefine a single point to make the function continuous. The first type of non-removable discontinuity is a jump discontinuity, which occurs when both one-sided limits exist and are finite, but they are not equal to each other. That means the function "jumps" from one value to another at $x=a$, so there is no way to pick a single value for $f(a)$ to make the two-sided limit exist. Jump discontinuities almost always occur in piecewise functions with different expressions on either side of $a$, in functions with absolute value rewritten as piecewise, or in step functions like the greatest integer function $\lfloor x\rfloor$. For example, $f(x)=\{\begin{array}{ll}2x+1& x<1\\ 2x-3& x\ge 1\end{array}$ has a jump discontinuity at $x=1$, where the left limit is $3$ and the right limit is $-1$.

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: $x^2-4=(x-2)(x+2)$, so 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 one-sided limits: Left limit: $$\underset{x\to 2^{-}}{\lim }\frac{-(x-2)(x+2)}{x-2}=\underset{x\to 2^{-}}{\lim }-(x+2)=-4$$ Right limit: $$\underset{x\to 2^{+}}{\lim }\frac{(x-2)(x+2)}{x-2}=\underset{x\to 2^{+}}{\lim }(x+2)=4$$
3. Check conditions: $g(2)$ is undefined, both one-sided limits exist and are finite, but ${\lim }_{x\to 2^{-}}g(x)\ne {\lim }_{x\to 2^{+}}g(x)$, so the two-sided limit does not exist.
4. This matches the definition of a jump discontinuity, a non-removable discontinuity.

Exam tip: Always evaluate both one-sided limits when working with piecewise functions or functions involving absolute value at the boundary point. Never assume the discontinuity is removable without checking that the two one-sided limits are equal.

4. Non-Removable Discontinuities: Infinite Discontinuities

The second common type of non-removable discontinuity is an infinite discontinuity, which occurs when one or both of the one-sided limits as $x\to a$ is infinite (either $+\infty$ or $-\infty$). These correspond to vertical asymptotes of the function, which occur when a factor of the denominator does not cancel with a matching factor in the numerator of a rational function. For example, $f(x)=\frac{1}{(x-2{)}^2}$ has an infinite discontinuity at $x=2$, since both one-sided limits are $+\infty$. $f(x)=\frac{1}{x-3}$ has an infinite discontinuity at $x=3$, with left limit $-\infty$ and right limit $+\infty$. In both cases, the limit does not exist (it is infinite, not finite), so the discontinuity cannot be removed by redefining $f(a)$. It is important to distinguish infinite discontinuities from jump discontinuities: jump discontinuities have finite one-sided limits, while infinite discontinuities have at least one infinite one-sided limit.

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: $$\underset{x\to -5}{\lim }\frac{x+5}{(x-5)(x+5)}=\underset{x\to -5}{\lim }\frac{1}{x-5}=-\frac{1}{10}$$ The limit exists and is finite, but $h(-5)$ is undefined, so $x=-5$ is a removable discontinuity.
3. Evaluate the one-sided limits at $x=5$: $$\underset{x\to 5^{-}}{\lim }\frac{1}{x-5}=-\infty \text{and}\underset{x\to 5^{+}}{\lim }\frac{1}{x-5}=+\infty$$
4. At least one one-sided limit is infinite, so $x=5$ is an infinite non-removable discontinuity.

Exam tip: Always cancel common factors first to separate removable discontinuities from vertical asymptotes (infinite discontinuities). A common student error is misclassifying a cancelled factor as an infinite discontinuity.

5. Oscillating Discontinuities

A less frequently tested but still AP-examinable type of non-removable discontinuity is the oscillating discontinuity, which occurs when 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. The classic example is $f(x)=\sin \left(\frac{1}{x}\right)$ at $x=0$. As $x\to 0$, $\frac{1}{x}$ goes to infinity, so $\sin (1/x)$ oscillates between $-1$ and $1$ infinitely many times, never approaching a single value. Because the limit does not exist, this is a non-removable discontinuity. This type can also occur with growing amplitude, such as $f(x)=\frac{1}{x}\sin \left(\frac{1}{x}\right)$ at $x=0$, but the classification remains the same. On the AP exam, this is most often tested in multiple-choice questions asking to identify discontinuity type from a description or graph.

Worked Example

The function $f(x)=\cos \left(\frac{2}{x+1}\right)$ 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 $\cos \left(\frac{2}{x+1}\right)$ oscillates between $-1$ and $1$ infinitely many times, never approaching a single finite limit, and does not approach $\pm \infty$.
3. Compare to definitions: 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).
4. This is an oscillating non-removable discontinuity.

Exam tip: If the function is a trigonometric function of $\frac{1}{x-a}$, it will always have an oscillating discontinuity at $x=a$, no need to overcomplicate the classification.

6. Common Pitfalls (and how to avoid them)

• Wrong move: Classifying $f(x)=\frac{(x-2)(x+1)}{x-2}$ 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: Classifying 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: Claiming that ${\lim }_{x\to 0}\sin (1/x)=0$, 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 $f(x)=\{\begin{array}{ll}{x}^2& x<1\\ 1& x=1\\ 3x-2& x>1\end{array}$, claiming $x=1$ has a jump discontinuity because $f(1)=1$ doesn't match... wait no, in this case left and right limits are both 1, so it's removable. Wrong move here is: claiming $x=1$ is jump because $f(1)$ is mismatched. Why: Students forget to check that one-sided limits are equal. 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: Stating 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. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

How many of the following have a non-removable discontinuity at the given point? I. $f(x)=\frac{x^2-4x}{x^2-16}$ at $x=4$ II. $g(x)=\frac{\lfloor x^2\rfloor }{x-2}$ at $x=2$ III. $h(x)=(x-3)e^{1/(x-3)}$ at $x=3$

(A) 0 (B) 1 (C) 2 (D) 3

Worked Solution: Evaluate each case one by one. For I: Factor numerator and denominator to get $\frac{x(x-4)}{(x-4)(x+4)}$, so $(x-4)$ cancels, and ${\lim }_{x\to 4}\frac{x}{x+4}=\frac{1}{2}$, which is finite, so I has a removable discontinuity. For II: ${\lim }_{x\to 2}\lfloor x^2\rfloor =4$, so ${\lim }_{x\to 2^{-}}\frac{4}{x-2}=-\infty$ and ${\lim }_{x\to 2^{+}}\frac{4}{x-2}=+\infty$, so this is an infinite non-removable discontinuity. For III: As $x\to 3^{-}$, $1/(x-3)\to -\infty$, so $e^{1/(x-3)}\to 0$, so the left limit is $0$, while the right limit as $x\to 3^{+}$ is $+\infty$, so the two-sided limit does not exist, meaning III is non-removable. Two of the three are non-removable, so the correct answer is (C).

Question 2 (Free Response)

Let $f(x)=\{\begin{array}{ll}\frac{x^3-4x^2+3x}{x^2-1}& x\ne 1\\ 1& x=1\end{array}$ (a) Find all points where $f(x)$ has discontinuities, and justify your answer. (b) Classify each discontinuity you found in part (a) as removable or non-removable. If non-removable, further classify as jump, infinite, or oscillating. (c) For any removable discontinuities, state what value $f(1)$ should be changed to to make $f(x)$ continuous at that point.

Worked Solution: (a) First, factor the denominator: $x^2-1=(x-1)(x+1)$, so $f(x)$ is undefined at $x=-1$, which means $x=-1$ is a discontinuity. At $x=1$, $f(x)$ is defined as $f(1)=1$. Factor the numerator: $x^3-4x^2+3x=x(x-1)(x-3)$, so for $x\ne \pm 1$, $f(x)=\frac{x(x-3)}{x+1}$. The limit as $x\to 1$ is $\frac{1(-2)}{2}=-1\ne f(1)=1$, so $x=1$ also violates the continuity condition. Discontinuities exist at $x=-1$ and $x=1$.

(b) For $x=1$: ${\lim }_{x\to 1}f(x)$ exists and is finite, so this is a removable discontinuity. For $x=-1$: Evaluate one-sided limits: ${\lim }_{x\to -1^{-}}\frac{x(x-3)}{x+1}=-\infty$ and ${\lim }_{x\to -1^{+}}\frac{x(x-3)}{x+1}=+\infty$, so this is a non-removable infinite discontinuity.

(c) For the removable discontinuity at $x=1$, we need ${\lim }_{x\to 1}f(x)=f(1)$, so $f(1)$ should be changed to $-1$ to make $f(x)$ continuous at $x=1$.

Question 3 (Application / Real-World Style)

A delivery company charges $\$2.50$ per pound for packages up to 2 pounds, and $\$1.50$ per pound for any weight over 2 pounds. There is no fixed base fee. Let $C(w)$ be the total cost (in dollars) for a package of weight $w$ pounds, $w>0$. Identify the location of any discontinuity of $C(w)$, classify its type, and explain what the discontinuity means in the context of the company's pricing structure.

Worked Solution: First, write the piecewise function for $C(w)$: $$C(w)=\{\begin{array}{ll}2.50w& 0<w\le 2\\ 1.50w& w>2\end{array}$$ The only candidate for discontinuity is the boundary at $w=2$. Evaluate one-sided limits: ${\lim }_{w\to 2^{-}}C(w)=2.50(2)=5.00$, ${\lim }_{w\to 2^{+}}C(w)=1.50(2)=3.00$. $C(2)=5.00$, which is defined, but the two one-sided limits are finite and not equal, so $C(w)$ has a non-removable jump discontinuity at $w=2$. In context, this means the total cost of a package just over 2 pounds is almost 2 dollars less than the total cost of a package exactly at 2 pounds, representing a sudden price drop at the 2 pound weight threshold.

8. Quick Reference Cheatsheet

9. What's Next

Mastering the classification of discontinuities is a critical prerequisite for the next topics in Unit 1: applying the Intermediate Value Theorem (IVT), which requires that a function is continuous on an interval to apply, so you need to identify discontinuities to check the IVT's continuity condition. Beyond Unit 1, this topic feeds into finding derivatives of piecewise functions at boundary points, where discontinuities mean the derivative does not exist, and into improper integrals, where you need to identify infinite discontinuities at bounds to know if the integral is improper. You will also use this skill when analyzing convergence of power series, where discontinuities in the limit function of a series can occur at the endpoints of the interval of convergence.

• Intermediate Value Theorem
• Derivatives of Piecewise Functions
• Improper Integrals
• Power Series Convergence