A one-sided limit is the value a function approaches as x approaches a point from only one direction, written lim x→a⁻ f(x) (from the left) or lim x→a⁺ f(x) (from the right). On the AP exam, the two-sided limit exists only when both one-sided limits exist and are equal.
A one-sided limit asks a narrower question than a regular limit. Instead of "what does f(x) approach as x gets close to a?", it asks "what does f(x) approach as x gets close to a from the left only (values less than a) or from the right only (values greater than a)?" The notation is lim x→a⁻ f(x) for the left-hand limit and lim x→a⁺ f(x) for the right-hand limit.
Here's the rule that makes everything else work: the two-sided limit lim x→a f(x) exists if and only if both one-sided limits exist and agree. Think of it like two people walking toward the same spot from opposite directions. If they end up at the same height, the limit exists. If they arrive at different heights, you have a jump, and the limit does not exist. That comparison is exactly how you classify discontinuities and check continuity for piecewise functions in Unit 1.
One-sided limits live in Unit 1: Limits and Continuity and do heavy lifting in Topic 1.13: Removing Discontinuities, supporting learning objective AP Calc 1.13.A (determine values of x or solve for parameters that make discontinuous functions continuous). Per EK LIM-2.C.1, a discontinuity is removable only if the limit at that point exists, and you verify that by checking that the left and right limits match. Per EK LIM-2.C.2, a piecewise function is continuous at a boundary only when the expression on one side of the boundary equals the expression on the other side at that point. That "set the left piece equal to the right piece and solve for the parameter" move is one of the most reliable problem types in Unit 1, and it's a one-sided limit comparison in disguise.
Keep studying AP Calculus Unit 1
Visual cheatsheet
view galleryLimit (Unit 1)
A two-sided limit is really two one-sided limits that happen to agree. Whenever an AP question asks whether lim x→a f(x) exists at a suspicious point, your actual job is to compute the left-hand and right-hand limits and compare them.
Continuity (Unit 1)
Continuity at a point requires three things, and the limit existing is the one that breaks most often. For piecewise functions, EK LIM-2.C.2 turns continuity into a one-sided limit equation. The piece on the left of the boundary must equal the piece on the right, and both must equal the function's value there.
Discontinuity (Unit 1)
One-sided limits are your diagnostic tool for classifying discontinuities. If the left and right limits agree but the function value is wrong or missing, it's removable. If they disagree, it's a jump, which cannot be removed by redefining one point. If one side blows up to infinity, you're looking at a vertical asymptote.
Differentiability (Unit 2)
The same left-versus-right logic returns when you check whether a derivative exists. A function with a corner (like |x| at x = 0) has different slopes coming from each side, so the derivative limit fails for the same reason a jump discontinuity fails. One-sided thinking from Unit 1 is the warm-up.
Multiple-choice questions love giving you a graph or a piecewise function and asking for lim x→2⁻ f(x), lim x→2⁺ f(x), or whether the full limit exists. Practice questions in this area hit the same moves repeatedly: use one-sided limits to identify what kind of discontinuity you're facing, explain why a jump discontinuity can't be removed (the one-sided limits disagree, so no single value fixes it), and solve for a parameter k that makes a piecewise function continuous by setting the left expression equal to the right expression at the boundary. On free-response questions, justifications about continuity at a point are expected to reference the limit from the left equaling the limit from the right, so write both one-sided limits explicitly rather than just asserting "the limit exists."
A two-sided limit (the plain lim x→a f(x)) requires the function to approach the same value from both directions. A one-sided limit only checks one direction, so it can exist even when the two-sided limit doesn't. At a jump discontinuity, both one-sided limits exist as perfectly good numbers, but because they're different numbers, the two-sided limit does not exist. Saying "the limit is 3 from the left" is not the same as saying "the limit is 3."
A one-sided limit measures what f(x) approaches as x approaches a point from just one direction, with x→a⁻ meaning from the left and x→a⁺ meaning from the right.
The two-sided limit exists if and only if the left-hand limit and the right-hand limit both exist and are equal.
One-sided limits can exist even when the overall limit does not, which is exactly what happens at a jump discontinuity.
A discontinuity is removable only when the one-sided limits agree, because then you can redefine the function's value at that point to equal the limit (EK LIM-2.C.1).
To make a piecewise function continuous at a boundary, set the expression from the left equal to the expression from the right at that x-value and solve for the unknown parameter (EK LIM-2.C.2).
In FRQ justifications, show both one-sided limits explicitly when arguing a function is or isn't continuous at a point.
One-sided limits describe what a function approaches as x approaches a point from a single direction, written lim x→a⁻ f(x) for the left side and lim x→a⁺ f(x) for the right side. They're the building blocks for deciding whether a full limit exists and whether a function is continuous.
No, not necessarily. The two-sided limit exists only when both one-sided limits exist and are equal. At a jump discontinuity, both one-sided limits exist as real numbers, but the overall limit does not exist because they disagree.
A regular (two-sided) limit checks both directions at once, while a one-sided limit only checks the approach from the left or the right. On the exam, you compute the two one-sided limits separately and compare them to determine whether the regular limit exists.
At the boundary where the pieces switch, set the value of the left expression equal to the value of the right expression and solve for the unknown parameter. This is learning objective 1.13.A, and it's one of the most common Unit 1 problem types.
No. A jump discontinuity means the left-hand and right-hand limits are different numbers, so no single redefined function value can match both. Only discontinuities where the one-sided limits agree (removable discontinuities) can be fixed by redefining one point.