The Meaning of a Limit

You’ll be able to

Interpret a limit as the value a function approaches, independent of the value at the point
Distinguish two-sided limits from one-sided limits and state when a limit fails to exist
Evaluate limits from a graph, a table, and by direct substitution

A limit is about approach, not arrival

The statement lim(x→a) f(x) = L means that as x gets arbitrarily close to a — from both sides — the outputs f(x) get arbitrarily close to L. Crucially, the limit says nothing about f(a) itself. The function need not even be defined at a; a limit describes the trend of the outputs near a, not the value at a. This separation between "where the function is heading" and "where it actually is" is the whole reason limits exist.

One-sided limits and existence

The left-hand limit lim(x→a⁻) f(x) watches x approach a from values below a; the right-hand limit lim(x→a⁺) f(x) watches x approach from above. The two-sided limit exists only when both one-sided limits exist and agree. If the left and right limits disagree — as at a jump — the two-sided limit does not exist (DNE). A limit also fails to exist when the outputs grow without bound or oscillate forever.

Evaluating: try substitution first

For a function built from polynomials, roots, exponentials, and trig — anything continuous at a — the limit is simply the value: lim(x→a) f(x) = f(a). So the first move is always direct substitution. Only when substitution produces an indeterminate form such as 0/0 do you need more work: factoring, rationalizing, or a special-limit trick. A finite nonzero-over-zero result instead signals a vertical asymptote, not a removable issue.

Existence of a two-sided 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 differ, the two-sided limit does not exist.

Two limits worth memorizing

lim(x→0) sin(x)/x = 1 · lim(x→0) (1 − cos x)/x = 0

These special trig limits appear constantly; direct substitution gives 0/0, but the true limits are 1 and 0.

Worked example

Evaluate lim(x→3) (x² − 9)/(x − 3).

1.Try direct substitution: (3² − 9)/(3 − 3) = 0/0, an indeterminate form — more work is needed.
2.Factor the numerator as a difference of squares: x² − 9 = (x − 3)(x + 3).
3.Cancel the common factor: (x − 3)(x + 3)/(x − 3) = x + 3, valid for all x ≠ 3.
4.Now substitute into the simplified expression: 3 + 3 = 6.

Answer: The limit is 6. The original function has a removable hole at x = 3, but the outputs approach 6.

Watch out

A 0/0 result does not mean the limit is 0 or that it fails to exist — it means the form is indeterminate and you must simplify. Many students stop at 0/0; that is exactly where the real work begins.

Checkpoint

A function has lim(x→2⁻) f(x) = 5 and lim(x→2⁺) f(x) = 5, but f(2) = 1. What is lim(x→2) f(x)?

Checkpoint

Evaluate lim(x→0) sin(5x)/x.

On the exam

On the AP exam, always try substitution first and read the form. Continuous-looking value → done. 0/0 → simplify (factor/rationalize/special limit). Nonzero/0 → infinite limit (asymptote). Naming the form tells you the technique.

Answer the 2 checkpoints as you read.