# Exploring types of discontinuities

> AP Calculus BC · AP CED Unit 1: Limits and Continuity
> Source: https://www.owlsprep.com/study/ap-calculus-bc-u1-exploring-types-of-discontinuities/

This guide covers classification of discontinuity types (removable, jump, infinite, oscillating) for AP Calculus BC. You'll learn to categorize discontinuities from graphs and algebra, per AP CED standards, with worked examples and exam tips.

**Prerequisites:** [Evaluating one-sided and two-sided limits](https://www.owlsprep.com/study/ap-calculus-bc-u1-limit-evaluation/); [Definition of continuity at a point](https://www.owlsprep.com/study/ap-calculus-bc-u1-continuity-definition/); Algebraic simplification of rational and piecewise functions

## Learning objectives

- Classify discontinuities as removable or non-removable
- Further classify non-removable discontinuities as jump, infinite, or oscillating
- Use one-sided and two-sided limits to justify discontinuity type
- Apply discontinuity classification to rational, piecewise, and trigonometric functions

## Overview of Discontinuity Classification

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.

**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.

## Removable Discontinuities

**Removable Discontinuity** — 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.

*Notation:* Also called a hole

*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 $f(x) = \frac{x^2 - 9}{x^2 - 2x - 3}$ 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.

> **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.

## Non-Removable Discontinuities: Jump

**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 $\lfloor x \rfloor$

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.

> **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.

## Non-Removable Discontinuities: Infinite

**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.

> **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.

## Non-Removable Discontinuities: Oscillating

**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 $\sin\left(\frac{1}{x}\right)$ 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 $f(x) = \cos\left(\frac{2}{x+1}\right)$ 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. 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.

> **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.

**Check your understanding**

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

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

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

   *Why:* I has a removable discontinuity, II and III are non-removable, so 2 of 3 are non-removable.

## Common pitfalls

- **Wrong:** Classify $f(x) = \frac{(x-2)(x+1)}{x-2}$ as having an infinite discontinuity at $x=2$, because the denominator is zero.
  - Why it fails: Students associate zero denominator with vertical asymptotes without factoring and checking for common factors.
  - Correct: Always factor numerator and denominator completely, cancel common factors, and evaluate the limit at the point before classifying.
- **Wrong:** Classify a jump discontinuity at $x=a$ as removable because $f(a)$ is undefined.
  - Why it fails: 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: 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:** Claim $\lim_{x \to 0} \sin(1/x) = 0$, so it is a removable discontinuity, because the function oscillates around zero.
  - Why it fails: Students mistake the midpoint of oscillation for the limit.
  - Correct: 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:** For $f(x) = \begin{cases} x^2 & x < 1 \\ 1 & x = 1 \\ 3x - 2 & x > 1 \end{cases}$, claim $x=1$ has a jump discontinuity because $f(1)=1$ doesn't match the expected value.
  - Why it fails: Students forget to check that one-sided limits are equal before classifying.
  - Correct: Always evaluate left and right limits first; if both are equal, the discontinuity is removable regardless of the value of $f(a)$.
- **Wrong:** State that any discontinuity where $f(a)$ is undefined is removable.
  - Why it fails: Students overgeneralize the case of cancelled common factors.
  - Correct: Only classify as removable if the two-sided limit at $a$ is finite, regardless of whether $f(a)$ is defined or not.

## 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 | 1. $f(a)$ defined; 2. $\lim_{x \to a} f(x)$ exists; 3. $\lim = f(a)$ | Discontinuity if any condition fails |

## 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.

- [Defining continuity at a point](https://www.owlsprep.com/study/ap-calculus-bc-u1-defining-continuity-at-a-point/)
- [Confirming continuity over an interval](https://www.owlsprep.com/study/ap-calculus-bc-u1-confirming-continuity-over-an-interval/)
- [Removing Discontinuities](https://www.owlsprep.com/study/ap-calculus-bc-u1-removing-discontinuities/)

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