AP Calculus AB and BC glossary

Bounded Function

A bounded function is one whose outputs all stay inside a fixed interval, so there is some number M that f(x) never exceeds in size anywhere on the domain. Bounded is a statement about size and nothing else: sin(1/x) never leaves the range from -1 to 1 and still has no limit at 0.

\exists\, M > 0 \ \text{with}\ |f(x)| \le M \ \text{for all } x \text{ in the domain}

The squeeze theorem runs on boundedness. Because $-1 \le \sin(1/x) \le 1$ for every $x \neq 0$ , multiplying by the non negative quantity $x^2$ traps the product between two curves that both go to $0$ .

-x^2 \le x^2 \sin\!\left(\frac{1}{x}\right) \le x^2

Being bounded does not mean the bounds get reached. The graph of $\arctan x$ stays strictly between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ and touches neither. Continuity on a closed, bounded interval $[a, b]$ is the stronger situation: there the extreme value theorem promises a bound and a point where the function actually attains it. Both hypotheses earn their place, and $\arctan x$ on $[0, \infty)$ shows why: the interval is closed, the function is continuous and bounded, and no maximum is ever attained.

The mistake

Reading bounded as convergent. A bounded function is free to oscillate forever, and so is a bounded sequence, so trapped between two numbers is never on its own a reason to claim a limit exists. For a sequence, bounded plus monotonic does force convergence, which is the Monotone Convergence Theorem. No such rule rescues a function at a finite point: $f(x) = x$ for $x < 0$ and $f(x) = x + 1$ for $x \ge 0$ is increasing and bounded on $(-1, 1)$ , with no limit at $0$ .

Appears in: Unit 1: Limits and Continuity