Derivative of bounded function when $x \to \infty$

Question

Can it be said for a differentiable bounded function $f(x)$ that if $\lim_{x \to \infty} f(x)$ exists then $\lim_{x \to \infty} f'(x)=0$ ?

This holds for functions like $\arctan(x)$ and arccot$(x)$ but will it always be true and can it be proved?

Small nitpick, you probably want to require the $f$ is differentiable otherwise there's trivial counterexamples. — , Nov 19 '17 at 16:27
@Mattos: in your example it is assumed that $\lim_{x \to \infty} f'(x)$ exists. So the hypotheses are different than here, so I would say it is not a duplicate. — Robert Lewis, Nov 19 '17 at 17:04
No. But yes if you also assume that $\lim_{x\to\infty}f'(x)$ exists. — David C. Ullrich, Nov 19 '17 at 17:24

score 3 · Answer 1 · answered Nov 19 '17 at 16:39

Nope.

Counterexample:

Consider, for $n \in \Bbb N$,

$f(x) = \dfrac{1}{x^2 + 1} \sin x^n = (x^2 + 1)^{-1} \sin x^n; \tag 1$

we have

$f'(x) = -(x^2 + 1)^{-2}(2x) \sin x^n + nx^{n - 1}(x^2 + 1)^{-1} \cos x^n$ $= -2x(x^2 + 1)^{-2} \sin x^n + nx^{n - 1}(x^2 + 1)^{-1}\cos x^n; \tag 2$

for $n \ge 4$, as $x \to \infty$,

$f(x) \to 0, \tag 3$

but for any $B, M \in \Bbb R$ there exists $x > B$ with

$f'(x) > M; \tag 4$

thus

$f'(x) \not \to 0 \; \text{as} \; x \to \infty; \tag 5$

we see in fact that $f'(x)$ may be arbitrarily large as $x \to \infty$, even though $f(x) \to 0$.

1 Answers1