One-Sided & Two-Sided Limits

You’ll be able to

Interpret a limit as the value a function approaches, independent of the value at the point
Evaluate left-hand and right-hand limits from graphs and piecewise rules
Determine when a two-sided limit exists by comparing the one-sided limits

A limit is about approach, not arrival

The statement lim(x→a) f(x) = L means that as x gets arbitrarily close to a — without ever equaling a — the outputs f(x) get arbitrarily close to L. The limit describes the trend of the function near a, and it deliberately ignores what happens at x = a. A function can have a hole, a jump, or a completely different value at a and still have a perfectly good limit there. That separation between "the value the function heads toward" and "the value it actually takes" is the whole engine of calculus.

Left and right: the two one-sided limits

You can approach a from either side. The left-hand limit lim(x→a⁻) f(x) uses only inputs slightly less than a; the right-hand limit lim(x→a⁺) f(x) uses only inputs slightly greater than a. These are the natural tools for piecewise functions, where a different rule applies on each side of the break. Read the superscript sign carefully: the minus means "from below on the number line", the plus means "from above".

When does the two-sided limit exist?

The full two-sided limit exists only when both one-sided limits exist and agree. In symbols, lim(x→a) f(x) = L requires lim(x→a⁻) f(x) = lim(x→a⁺) f(x) = L. If the left side heads toward one number and the right side heads toward a different number, the function jumps and the two-sided limit does not exist (DNE). This single test settles most "does the limit exist?" questions.

Existence of a limit

lim(x→a) f(x) = L ⟺ lim(x→a⁻) f(x) = L AND lim(x→a⁺) f(x) = L

Both one-sided limits must exist and be equal. If they disagree, the two-sided limit does not exist.

Worked example

Let f(x) = x + 1 for x < 2 and f(x) = 5 − x for x ≥ 2. Find lim(x→2⁻) f(x), lim(x→2⁺) f(x), and lim(x→2) f(x).

1.From the left (x < 2) use the rule f(x) = x + 1, so lim(x→2⁻) f(x) = 2 + 1 = 3.
2.From the right (x ≥ 2) use the rule f(x) = 5 − x, so lim(x→2⁺) f(x) = 5 − 2 = 3.
3.Compare the one-sided limits: both equal 3, so they agree.
4.Because the left and right limits match, the two-sided limit exists and equals that common value.

Answer: lim(x→2⁻) f(x) = 3, lim(x→2⁺) f(x) = 3, so lim(x→2) f(x) = 3 — even though the two pieces are given by different formulas.

Watch out

The value f(a) has nothing to do with whether the limit exists. A function can be defined as f(2) = 100 yet still have lim(x→2) f(x) = 3. Never plug the point in and assume that number is the limit — read what the graph approaches.

Checkpoint

A function has lim(x→3⁻) f(x) = 4 and lim(x→3⁺) f(x) = 7. What is lim(x→3) f(x)?

Checkpoint

For g(x) = (x² − 9)/(x − 3), what is lim(x→3) g(x)?

On the exam

For piecewise functions, always compute the two one-sided limits separately using the correct branch, then compare. Writing "lim⁻ = lim⁺ = L, therefore the limit exists" is exactly the justification AP readers want to see.

Answer the 2 checkpoints as you read.