A jump discontinuity occurs at a point where both one-sided limits of a function exist and are finite but are not equal to each other, so the overall limit does not exist and the graph makes an abrupt vertical leap, most often at the boundary of a piecewise-defined function.
A jump discontinuity is what you get when a function's left-hand limit and right-hand limit at a point both exist and are finite, but they disagree. The graph literally jumps from one y-value to another with no connection between the pieces. Because the two one-sided limits don't match, the two-sided limit does not exist at that point, and that's exactly why the function fails the definition of continuity there.
The classic place to find one is at the boundary of a piecewise-defined function, where one rule hands off to another. If the two rules don't agree at the boundary x-value, the graph leaps. For example, f(x) = |x|/x equals -1 for all negative x and +1 for all positive x, so at x = 0 the graph jumps from -1 up to 1. The CED lists jump discontinuities alongside removable discontinuities and vertical asymptotes as the three discontinuity types you need to recognize (EK LIM-2.A.1).
Jump discontinuities live in Unit 1: Limits and Continuity, specifically Topic 1.10 (Exploring Types of Discontinuities) and Topic 1.13 (Removing Discontinuities). Learning objective 1.10.A asks you to justify conclusions about continuity using the actual definition, which means checking three things at a point. Does f(a) exist? Does the limit exist? Are they equal? A jump discontinuity fails at step two, because mismatched one-sided limits mean no limit at all. That makes 'jump' the fastest classification to spot once you compute both one-sided limits.
It also matters for Topic 1.13 in a backwards way. EK LIM-2.C.1 says a discontinuity can only be removed if the limit exists there. A jump fails that condition, so a true jump discontinuity is not removable. But here's the twist the exam loves. When a piecewise function has an unknown parameter, you can often choose the parameter so the two sides match (EK LIM-2.C.2), which prevents the jump from existing in the first place. Continuity is also the gateway concept for everything later, since theorems like IVT and the definition of the derivative assume it.
Keep studying AP Calculus Unit 1
Visual cheatsheet
view galleryOne-sided Limits (Unit 1)
One-sided limits are the diagnostic tool for jumps. If the limit from the left and the limit from the right both exist but aren't equal, you have a jump. If they're equal, you don't. Every jump-discontinuity problem is secretly a one-sided limits problem.
Removable Discontinuity (Unit 1)
These are the two non-asymptote discontinuity types, and the limit is what separates them. A removable discontinuity has a limit that exists (just a hole or a misplaced point), so you can patch it by redefining f(a). A jump has no limit, so no single redefinition can fix it.
Piecewise Function (Unit 1)
Piecewise boundaries are where jumps are born. The exam's favorite version asks you to solve for a parameter (like k) that makes the two pieces agree at the boundary, which is exactly EK LIM-2.C.2. You're literally solving 'left-hand limit = right-hand limit' to eliminate a potential jump.
Vertical Asymptote (Unit 1)
The third discontinuity type in EK LIM-2.A.1. The difference is whether the function stays finite. At a jump, both one-sided limits are real numbers. At a vertical asymptote, like f(x) = 1/x at x = 0, at least one one-sided limit is infinite. Finite leap means jump, blow-up means asymptote.
Jump discontinuities show up most often in multiple-choice classification questions. A stem gives you a function, a graph, or a piecewise definition and asks which type of discontinuity occurs at a specific x-value. Your job is to compute or read both one-sided limits and sort the point into one of three bins (removable, jump, or vertical asymptote). Watch for trap answer choices, like 1/x at x = 0, which is a vertical asymptote and not a jump because the one-sided limits are infinite. Also watch for functions like |x|, which is perfectly continuous at x = 0 even though the graph has a sharp corner.
The other big format is the parameter problem tied to Topic 1.13. You're given a piecewise function with an unknown constant and asked for the value that makes the function continuous. Set the one-sided limits at the boundary equal to each other (and to the function's value) and solve. When you justify continuity in free-response work, write out the full definition: the limit exists, f(a) exists, and they're equal. Saying 'the pieces meet' won't earn the justification point.
Both break continuity without blowing up to infinity, so they look like siblings, but the limit tells them apart. At a removable discontinuity, the two-sided limit exists and the graph just has a hole (or a point floating in the wrong place), so redefining f(a) to equal the limit fixes it. At a jump, the one-sided limits exist but disagree, the two-sided limit does not exist, and no single value of f(a) can connect the two pieces. Quick test: if you can fill the gap with one dot, it's removable; if the graph leaps, it's a jump.
A jump discontinuity happens when the left-hand and right-hand limits at a point both exist and are finite but are not equal, so the two-sided limit does not exist.
Because the limit does not exist at a jump, a true jump discontinuity cannot be removed by redefining the function's value at that point (EK LIM-2.C.1).
Jumps almost always appear at the boundary of a piecewise-defined function where the two rules give different values.
To make a piecewise function with a parameter continuous, set the expressions on both sides of the boundary equal at that x-value and solve (EK LIM-2.C.2).
The three discontinuity types in the CED are removable, jump, and vertical asymptote, and you sort between them by checking whether the one-sided limits exist, match, or are infinite (EK LIM-2.A.1).
f(x) = |x| is continuous at x = 0 even though it has a sharp corner; a corner breaks differentiability, not continuity.
It's a point where a function's left-hand and right-hand limits both exist and are finite but are not equal, so the graph makes an abrupt vertical leap. It's one of the three discontinuity types listed in EK LIM-2.A.1 in Unit 1, along with removable discontinuities and vertical asymptotes.
No. Removing a discontinuity requires the limit to exist at that point, and at a jump the two-sided limit does not exist because the one-sided limits disagree. You can only 'fix' a jump in a parameter problem, by choosing the parameter so the jump never forms in the first place.
At a removable discontinuity the limit exists and there's just a hole you can fill by redefining f(a). At a jump, the one-sided limits are finite but unequal, so the limit doesn't exist and no single point can patch the gap.
No. The absolute value function is continuous everywhere, including x = 0, where both one-sided limits and f(0) all equal 0. The sharp corner means it's not differentiable there, but continuity and differentiability are different tests. A function with a jump is |x|/x, which leaps from -1 to 1 at x = 0.
No, it's a discontinuity due to a vertical asymptote. The one-sided limits at x = 0 are negative infinity and positive infinity, and a jump requires both one-sided limits to be finite. This exact distinction is a common multiple-choice trap.