A removable discontinuity is a point where a function is discontinuous but the limit still exists, leaving a single "hole" in the graph; per EK LIM-2.C.1, you can remove it by defining or redefining f(a) to equal the limit as x approaches a.
A removable discontinuity is the friendliest kind of break a graph can have. The function fails to be continuous at one point, but the limit from both sides still exists and agrees. Visually, it's a hole. Either the function has no value there at all, or the value is sitting somewhere off the curve (a hole with a stray dot above or below it). Everything about the function's behavior near that point is fine; only the value AT the point is wrong or missing.
The "removable" part is literal. EK LIM-2.C.1 says that if the limit exists at a discontinuity, you can remove it by defining or redefining f(a) to equal that limit. The classic example is a rational function like f(x) = (x² - 4)/(x - 2). At x = 2 you get 0/0, but factoring and canceling (x - 2) leaves x + 2, so the limit is 4. Define f(2) = 4 and the hole is patched. Compare that to jump discontinuities and vertical asymptotes (the other types listed in EK LIM-2.A.1), where the limit doesn't exist and no single value can fix the break.
Removable discontinuities live in Unit 1 (Limits and Continuity), specifically Topic 1.10 and Topic 1.13. They support two learning objectives. LO 1.10.A asks you to justify conclusions about continuity using the three-part definition (f(a) exists, the limit exists, and they're equal), and a removable discontinuity is exactly the case where the limit exists but the other conditions break down. LO 1.13.A asks you to solve for values or parameters that make a discontinuous function continuous, which is really just asking you to remove a removable discontinuity algebraically. This is also where the famous "find k so the piecewise function is continuous" problem comes from (EK LIM-2.C.2). The concept echoes later in Unit 6, where improper integrals (Topic 6.13) use the same core move of replacing a problem point with a limit.
Keep studying AP Calculus Unit 1
Visual cheatsheet
view galleryContinuity at a Point (Unit 1)
A removable discontinuity is what you get when exactly one piece of the continuity definition fails. The limit exists, but f(a) either doesn't exist or doesn't match it. Knowing which condition breaks is how you justify your answer on the exam.
One-sided Limits (Unit 1)
Removable means the left-hand and right-hand limits agree. If they disagree, you have a jump instead, and no amount of redefining f(a) can save it. Checking one-sided limits is the diagnostic test for which type of discontinuity you're looking at.
Rational Functions and Holes (Unit 1)
Most removable discontinuities you'll see come from rational functions where a factor cancels from the top and bottom. A canceled factor makes a hole; a factor that survives in the denominator makes a vertical asymptote. Same function, two very different discontinuities.
Improper Integrals (Unit 6)
Topic 6.13 deals with integrands that are unbounded or undefined somewhere in the interval, and the fix is the same instinct as Unit 1. You replace the problem point with a limit of a definite integral. If you internalized removable discontinuities, improper integrals feel familiar.
On multiple choice, you'll be asked to classify a discontinuity (removable, jump, or asymptote) from a graph, a formula, or a limit statement, exactly like the question "which type of discontinuity occurs when a function has a 'hole'?" You'll also see the algebra version, where you factor a rational expression, cancel the shared factor, and evaluate the limit at the hole. The piecewise version shows up on both MCQs and FRQs in the form "find the value of k that makes f continuous," which means setting the two pieces equal at the boundary (EK LIM-2.C.2) and solving. On free response, expect to justify continuity using the definition. The high-scoring justification names all three conditions and states which one fails, not just "there's a hole."
Both are breaks in a graph, but the limit behavior is completely different. At a removable discontinuity, the left and right limits agree, so the overall limit exists and you can patch the hole by redefining one point. At a jump discontinuity, the one-sided limits exist but disagree, so the limit does not exist and nothing you do to f(a) can fix it. Quick test: if you can fill in one dot and make the curve unbroken, it's removable. If the graph leaps to a different height, it's a jump.
A removable discontinuity is a hole in the graph where the limit exists but the function value is missing or doesn't match the limit.
You remove it by defining or redefining f(a) to equal the limit of f(x) as x approaches a (EK LIM-2.C.1).
In rational functions, a factor that cancels from numerator and denominator creates a removable discontinuity, while a factor left in the denominator creates a vertical asymptote.
For a piecewise function to be continuous at a boundary, both pieces must equal each other and the function's value there, which is the setup behind every 'solve for k' problem.
If the one-sided limits disagree, the discontinuity is a jump, not removable, and it cannot be fixed by redefining one point.
The same limit-based patching idea returns in Unit 6, where improper integrals handle unbounded integrands using limits of definite integrals.
It's a point where a function is discontinuous but the limit still exists, which shows up as a hole in the graph. Per EK LIM-2.C.1, you can remove it by setting f(a) equal to the limit as x approaches a.
Yes, and that's the defining feature. The left and right limits agree, so the overall limit exists. The function just has the wrong value (or no value) at that one point.
At a removable discontinuity the limit exists and is finite, so one redefined point fixes everything. At a vertical asymptote the function blows up to infinity, the limit doesn't exist as a finite number, and no redefinition can repair it. In a rational function, a canceled factor gives a hole while a surviving denominator factor gives an asymptote.
No, |x| is continuous everywhere, including x = 0. It has a sharp corner there, which matters for differentiability in Unit 2, but continuity and differentiability are different questions. This is a common trap in discontinuity-classification MCQs.
Set the two expressions equal at the boundary point and solve for k. EK LIM-2.C.2 requires both pieces to give the same value at the boundary, and that value must equal f at the boundary, so matching the pieces is the whole game.