5

I want to show that if $f:\mathbb R\longrightarrow \mathbb R$ (a derivable function) is bounded and s.t. $$\lim_{x\to \infty }f'(x)=0,$$ then $f$ has a limit in $+\infty $.

I tried as follow:

If $f$ doesn't reach his supremum (let denote it $\ell\in\mathbb R$), then I can construct a sequence $(x_n)_n$ s.t. $$\lim_{n\to \infty }f(x_n)=\ell.$$

But I can't do better. Any idea ?

MSE
  • 3,153
  • you're on the wrong track, $f$ does not necessarly converge towards it's supremum, it could oscillate, think of $x\mapsto sin(\frac 1 x)$ – marmouset Sep 08 '16 at 08:35
  • pick $(x_n)$ with $lim x_n =+\infty$, try to prove that $f(x_n)$ is Cauchy. – marmouset Sep 08 '16 at 08:37
  • I can see the intuition you were going for but in some cases, the derivative tends to 0 very very slowly; enough that the original function won't converge, even as you take x to infinity. – gowrath Sep 08 '16 at 09:18

3 Answers3

8

It's false. Counterexample: any $g\in C^1(\Bbb R)$ such that $g(x)=\sin\ln x$ for all $x>1$.

5

I don't think that the claim is true.

Let $f(x) = \sin (\sqrt x)$. Then

$$ f'(x) = \frac1{2\sqrt x} \cos (\sqrt x) \rightarrow 0 \ \mathrm{as} \ x\rightarrow\infty, $$

and $f$ is bounded, but $\lim\limits_{x\rightarrow\infty} f(x)$ does not exist.

This has a problem that $f$ is not differentiable everywhere, e. g. at $x=0$. To avoid this problem, we modify $f$ as follows:

$$ g(x) = \begin{cases} \sin (\sqrt x) &\mbox{if} \ x\geq (\pi/2)^2\\ 1 &\mbox{if} \ x<(\pi/2)^2 \end{cases} $$

Sungjin Kim
  • 20,102
2

False. A counterexample is $$ f(x)=\sin\left(\sqrt[3]{x}\right) $$ where $$ f'(x)=\frac{\cos\left(\sqrt[3]{x}\right)}{3\sqrt[3]{x^2}} $$

The function above is differentiable everywhere, however its derivative at $x=0$ is $+\infty$. Although the main point of the question is the behavior near $\infty$, we can adjust the example $$ f(x)=\sin\left(\sqrt[3]{x^2+1}\right) $$ where $$ f'(x)=\frac{2x\cos\left(\sqrt[3]{x^2+1}\right)}{3\sqrt[3]{x^4+2x^2+1}} $$

robjohn
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