Limits

A limit asks what a function is heading toward as x closes in on a number, not what happens when it gets there.

Explore: sneak up on a point

Pick a function, then drag the slider to move the blue point (from the left) and the amber point (from the right) toward the dashed line. Watch the outputs. Predict the limit before you reveal it.

The idea in one sentence

limx→a f(x) = L means f(x) gets as close to L as you like by taking x close enough to a (on both sides), whether or not x = a is allowed.

Two consequences you just saw. First, f(a) can be undefined or different from L, because the limit ignores the point itself. Second, x can approach a from the left (written x→a−) or from the right (x→a+), and the two sides get a vote.

When does a limit exist?

limx→a f(x) exists exactly when the left-hand limit and the right-hand limit both exist and are equal to the same finite number L.

It fails in three common ways, and you met each in the explorer: a jump (the sides disagree), unbounded behavior (outputs grow without bound, like 1/(x−1)), and oscillation (outputs swing forever, like sin(1/x) near 0). On the AP exam, writing “= ∞” describes why a limit fails to exist; it is still not a number.

Techniques for computing limits

Always try direct substitution first. If you get a real number, you’re done (for continuous functions). If you get 0/0, the expression needs rewriting.

Factor and cancel
limx→3 (x² − 9)/(x − 3) = lim (x + 3) = 6
Multiply by the conjugate
limx→0 (√(x+4) − 2)/x = lim 1/(√(x+4) + 2) = 1/4
Common denominator
limx→0 (1/(x+1) − 1)/x = lim −1/(x+1) = −1
Special trig limits
limx→0 sin x / x = 1 and limx→0 (1 − cos x)/x = 0
Squeeze theorem
If g(x) ≤ f(x) ≤ h(x) near a and g and h both approach L, then f approaches L. Example: x² sin(1/x) is squeezed between −x² and x², so its limit at 0 is 0.
Limits at infinity
Divide by the highest power: limx→∞ (3x² + 1)/(x² − 4) = lim (3 + 1/x²)/(1 − 4/x²) = 3

Common mistakes

Assuming the limit equals f(a)
The limit describes nearby behavior. In the explorer, (x² − 1)/(x − 1) is undefined at 1 but its limit is 2.
Treating 0/0 as an answer
0/0 is not 0, 1, or “undefined, so no limit.” It signals that you need to factor, use a conjugate, or simplify first.
Wrong: (x² − 9)/(x − 3) at 3 gives 0/0, so the limit is 0. Right: factor to x + 3, so the limit is 6.
Ignoring one side
Check both. If the left and right limits differ, the limit does not exist. Do not average them.
If f(x) = 2x for x < 1 and x + 2 for x ≥ 1, the sides give 2 and 3. The limit at 1 is not 2.5. It does not exist.
Calling “= ∞” a limit that exists
Writing lim = ∞ describes unbounded growth. It is not a real-number limit, so say the limit does not exist (or is infinite), depending on what the question asks.
As x → 0, 1/x² grows without bound, so lim = ∞. No real number is approached.
Dropping “lim” too early
Keep writing limx→a on every line until the step where you substitute. AP graders look for it.
Wrong: (x² − 9)/(x − 3) = x + 3 = 6. Right: limx→3 (x² − 9)/(x − 3) = limx→3 (x + 3) = 6.