A right-hand limit, written lim x→a⁺ f(x), is the value a function approaches as x approaches a from values greater than a (from the right side). On the AP exam, you compare it with the left-hand limit to test continuity and, in Topic 2.4, to decide whether a derivative exists.
A right-hand limit asks one specific question: as x slides toward a from the right side of the number line (values bigger than a), what y-value is f(x) heading toward? The notation is lim x→a⁺ f(x), and that little plus sign means "approaching from values greater than a," not "positive numbers."
The two-sided limit exists only when the right-hand limit and the left-hand limit agree. That single idea powers two big AP skills. First, checking continuity, especially for piecewise functions at the point where the rule changes. Second, checking differentiability in Topic 2.4. There, you apply one-sided limits not to f(x) itself but to the difference quotient. For f(x) = |x| at x = 0, the right-hand limit of the difference quotient is 1 and the left-hand limit is -1. They disagree, so f'(0) doesn't exist even though the function is perfectly continuous there. That's the corner.
This term lives in Unit 2 (Differentiation: Definition and Fundamental Properties), specifically Topic 2.4, and supports learning objective 2.4.A: explain the relationship between differentiability and continuity. The essential knowledge spells out exactly where right-hand limits earn their keep. A continuous function can still fail to be differentiable, and one way that happens is when the left-hand and right-hand limits of the difference quotient are not equal, as in f(x) = |x| at x = 0. In other words, the right-hand limit is the tool that lets you prove a corner exists. If you can compute one-sided limits of a difference quotient and compare them, you can explain why differentiability is a stricter condition than continuity, which is the whole point of Topic 2.4.
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view galleryLeft-hand limit (Units 1-2)
These are two halves of one idea. The two-sided limit exists only when the left-hand and right-hand limits match, and a derivative exists only when the left-hand and right-hand limits of the difference quotient match. Almost every right-hand limit problem on the exam is secretly a comparison problem.
Piecewise function (Unit 1)
Piecewise functions are where one-sided limits actually get computed. At the breakpoint, you plug into the right-side rule for the right-hand limit and the left-side rule for the left-hand limit. If the pieces disagree, the function jumps; if the slopes disagree, the derivative dies.
Closed Interval (Unit 1)
At the left endpoint a of a closed interval [a, b], there's nothing to the left of a inside the domain, so continuity there is defined using only the right-hand limit. Endpoints are the one place where a one-sided limit does the whole job by itself.
Differentiability and continuity (Unit 2)
Topic 2.4's headline result, differentiable implies continuous but not the other way around, is proven by counterexample using right-hand limits. f(x) = |x| at x = 0 is continuous, but the right-hand limit of its difference quotient is 1 while the left-hand limit is -1, so no derivative exists.
Right-hand limits show up two ways. In limits-and-continuity multiple choice, you'll read lim x→a⁺ off a graph or compute it from the correct piece of a piecewise function. In Unit 2, the stem changes. Practice questions ask things like "For f(x) = |x|, what is the right-hand limit of the difference quotient at x = 0?" (answer: 1) or "why does the derivative not exist at x = 0?" The move you must make is the same every time. Compute the right-hand limit, compute the left-hand limit, compare. Equal means the limit (or derivative) exists; unequal means it doesn't, and you should say why in your justification. Absolute value functions like f(x) = |x - 3| + 2 are the classic trap, continuous everywhere but not differentiable at the corner, and one-sided limits of the difference quotient are how you prove it.
The superscript trips people up. lim x→a⁺ means x approaches a from values greater than a (from the right on the number line), and lim x→a⁻ means from values less than a. The + and - describe which side of a you're coming from, not whether x is positive or negative. So lim x→-2⁺ still means approaching -2 from the right, through values like -1.9 and -1.99.
A right-hand limit, lim x→a⁺ f(x), is the value f(x) approaches as x approaches a from values greater than a.
The plus sign in x→a⁺ means "from the right side of a," not "positive x-values."
A two-sided limit exists only if the right-hand limit and left-hand limit are equal.
In Topic 2.4, you apply right-hand limits to the difference quotient. If the right-hand and left-hand limits of the difference quotient disagree, the derivative does not exist at that point.
For f(x) = |x| at x = 0, the right-hand limit of the difference quotient is 1 and the left-hand limit is -1, which is exactly why the corner has no derivative.
For piecewise functions, compute the right-hand limit using the rule that applies for x-values greater than the breakpoint.
It's the value a function approaches as x approaches a point from values greater than that point, written lim x→a⁺ f(x). You compare it with the left-hand limit to decide whether the full limit exists.
No. The plus sign means you're approaching a from the right side, through values greater than a. For example, lim x→-2⁺ means approaching -2 through values like -1.9 and -1.99, all of which are negative.
A two-sided limit requires the function to approach the same value from both directions. The right-hand limit only checks the approach from values greater than a. The two-sided limit exists if and only if both one-sided limits exist and are equal.
Not necessarily. Equal one-sided limits of f(x) give you the limit (and possibly continuity), but differentiability needs the one-sided limits of the difference quotient to match too. f(x) = |x| at x = 0 has matching limits of the function but mismatched difference-quotient limits (1 from the right, -1 from the left), so it's continuous but not differentiable.
It's 1, because for x > 0 the difference quotient is |x|/x = 1. The left-hand limit is -1, and since 1 ≠ -1, f'(0) does not exist. This is the standard Topic 2.4 example of continuity without differentiability.
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