AP Calculus AB and BC glossary

Limit of a composite function

Also called: Limit of a composition

To find the limit of a composite function, take the inner limit first, call it L, then apply the outer function to L. That step is guaranteed valid when the outer function is continuous at L. If it is discontinuous at L, the composite limit can differ from its value there or fail to exist.

\lim_{x \to a} g(x) = L \text{ and } f \text{ continuous at } L \implies \lim_{x \to a} f(g(x)) = f(L)

Worked the easy way, $\lim_{x \to 2} \ln(x^2 - 3)$ has inner limit $4 - 3 = 1$ , and $\ln$ is continuous at $1$ , so the answer is $\ln 1 = 0$ . Almost every composite limit built from polynomials, roots, exponentials, and trigonometric functions works like this, which is why the rule usually feels invisible.

The rule becomes visible on graph and table questions where the outer function has a removable discontinuity. Suppose $\lim_{x \to 1} g(x) = 2$ while $f$ satisfies $\lim_{u \to 2} f(u) = 3$ and $f(2) = 5$ . If $g(x)$ stays away from $2$ near $x = 1$ , then $\lim_{x \to 1} f(g(x)) = 3$ , following the limit of $f$ rather than its value. If instead $g$ hits $2$ exactly at points arbitrarily close to $1$ while also taking values other than $2$ arbitrarily close to $1$ , the composite bounces between $3$ and $5$ and the limit does not exist. If $g$ is simply equal to $2$ on a whole interval around $x = 1$ , the composite is constantly $f(2) = 5$ , so the limit is $5$ .

The mistake

Substituting the inner limit into the outer function without checking continuity. When $f$ has a hole at $L$ and $g$ avoids the value $L$ near $a$ , the answer is $\lim_{u \to L} f(u)$ , not $f(L)$ , so reading $f(L)$ off the graph gives the wrong number. On any composite limit stated through graphs or tables, find the inner limit, then check whether $g$ stays away from that value near $a$ . If it does, read what $f$ approaches there rather than what $f$ equals there.

Appears in: Unit 1: Limits and Continuity