AP® Calculus AB review sheet from Aim for Five (aimforfive.com/calc-ab/units/1/1-10)
Unit 1 · Topic 1.10
1.10 Exploring Types of Discontinuities
A discontinuity is a point where a graph breaks. There are three main kinds: removable (a hole), jump (the two sides end at different heights) and infinite (a vertical asymptote). You classify each one by looking at the limits there.
Key terms
- removable discontinuity
- jump discontinuity
- infinite discontinuity
- hole
- vertical asymptote
The three types
Every break falls into one of these three types, and each one has its own limit pattern. (Oscillation, like sin(1/x) at 0, is a fourth way to break continuity, but the exam focuses on these three.)
| Type | What the graph looks like | What the limits do |
|---|---|---|
| Removable | A hole, maybe with a dot somewhere else | lim (x→c) f(x) exists, but f(c) is missing or different |
| Jump | The graph leaps from one height to another | Left and right limits exist but are not equal |
| Infinite | The graph shoots up or down along x = c | At least one one-sided limit is ∞ or −∞ |
Finding discontinuities in rational functions
A rational function (a polynomial over a polynomial) is only discontinuous where its denominator is 0. At each such x = c, factor the top and bottom and cancel common factors. Then:
- If (x − c) cancels completely from the denominator, the limit exists: removable discontinuity (a hole).
- If a factor of (x − c) is left in the denominator after canceling, the function is unbounded near c: infinite discontinuity (vertical asymptote).
- Example: (x² − 1)/(x − 1) has a hole at x = 1, because the factor cancels and the limit is 2. But 1/(x − 1)² has a vertical asymptote at x = 1, because nothing cancels and the outputs grow without bound.
Finding discontinuities in piecewise functions
Each piece is usually a continuous formula, so the only suspects are the break points where the rule switches. Compute the left-hand limit with the left piece and the right-hand limit with the right piece. If they differ, it's a jump. If they match but f(c) is different or missing, it's removable.
Absolute value expressions like |x − 3| / (x − 3) also create jumps, since they equal −1 on one side and 1 on the other.
Removable vs. nonremovable
Removable discontinuities get their name because you could “remove” the break by filling in or moving a single point (1.13). The limit at c exists, so there's an obvious value to fill in.
Jump and infinite discontinuities are called nonremovable. No single point fixes them, because the limit at c doesn't exist. On a graph, a jump looks like a step and an infinite discontinuity looks like the curve escaping up or down a vertical line.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Classifying in a rational function
Find and classify all discontinuities of f(x) = (x² − 4) / (x² − x − 2).
Show the solutionHide the solution
- Step 1: The denominator is 0 when x² − x − 2 = (x − 2)(x + 1) = 0, so x = 2 or x = −1.
- Step 2: Factor the top: x² − 4 = (x − 2)(x + 2). For x ≠ 2, f(x) = (x + 2) / (x + 1).
- Step 3: At x = 2: (x − 2) canceled, and lim (x→2) f(x) = 4/3. But f(2) is undefined. Removable.
- Step 4: At x = −1: the factor (x + 1) is still in the denominator, while the top approaches 1. The outputs are unbounded. Infinite.
Answer: Removable discontinuity (hole) at x = 2, at the point (2, 4/3); infinite discontinuity (vertical asymptote) at x = −1.
- Example 2
Classifying in a piecewise function
g(x) = x + 1 for x < 1, and g(x) = 5 − x for x ≥ 1. Is there a discontinuity at x = 1? If so, what type?
Show the solutionHide the solution
- Step 1: Left-hand limit uses x + 1: lim (x→1⁻) g(x) = 2.
- Step 2: Right-hand limit uses 5 − x: lim (x→1⁺) g(x) = 4.
- Step 3: Both one-sided limits exist but are different.
Answer: Yes. g has a jump discontinuity at x = 1 (from height 2 to height 4).
- Example 3
Trap: a dot in the wrong place is not a jump
f(x) = x² for x ≠ 3, and f(3) = 1. A student says f has a jump discontinuity at x = 3 because the graph “jumps” down to the dot. Classify the discontinuity correctly.
Show the solutionHide the solution
- Step 1: Left-hand limit: x² → 9. Right-hand limit: x² → 9. Both sides agree, so lim (x→3) f(x) = 9.
- Step 2: f(3) = 1, which is not 9, so f is not continuous at 3.
- Step 3: A jump needs the left and right limits to be different. Here they're the same; only the single point is out of place.
Answer: Removable discontinuity at x = 3: the limit is 9 but f(3) = 1.
Common mistakes
- Calling every zero of the denominator a vertical asymptote. Factor and cancel first; a canceled factor gives a hole.
- Calling a hole “not a discontinuity” because the limit exists. It is still a discontinuity, just a removable one.
- Classifying a break in a piecewise function without computing both one-sided limits.
On the exam
- Expect multiple-choice questions that ask which type of discontinuity a function has at a point, from a formula or a graph.
- Removable discontinuities set up the next topics: continuity at a point (1.11) and redefining a function to remove a break (1.13).
Connected topics
Videos
Check yourself
4 questions on 1.10 Exploring Types of Discontinuities. Pick an answer to see if you got it, and why.
The function f is given by f(x) = (x² − 2x − 3)/(x² − 9). Which of the following describes the discontinuities of f ?
Let f(x) = (x² − 4)/|x − 2|. Which of the following describes the behavior of f at x = 2 ?
Which of the following functions has a jump discontinuity at x = 0 ?
A function h has a removable discontinuity at x = 3, and lim (x→3) h(x) = −2. Which of the following could be h(x)?
0 of 4 answered