If $f$ is differentiable in $(1,\infty )$ and $\lim_{x\to\infty }f'(x)=L<\infty $ then $\lim_{x\to\infty }f(x)=l\le\infty$?

Question

I need to prove or disprove this:

If $f$ is differentiable on $(1,\infty)$ and $\lim\limits_{x\to\infty}f'(x)=L\lt \infty$, then $\lim\limits_{x\to\infty}f(x)=\ell\leq\infty$.

After I didn't find any function to disprove with,

I started to think that if $\lim\limits_{x\to\infty }f'(x)=L<\infty $ so $f'$ is bounded and therfore $f$ is uniformly continuous but it doesn't mean that $\lim\limits_{x\to\infty }f(x)=l\leq\infty$, for example : $\sin x$ which is uniformly continuous, but it's limit when $x\to\infty$ does not exist.

What do you think?

Thank you.

Zarrax · Accepted Answer · 2011-06-17T16:01:49.780

4

Look at $f(x) = x^{1 \over 2} \sin(x^{1 \over 3})$. Then $f'(x) = {1 \over 2} x^{-{1 \over 2}}\sin(x^{1 \over 3}) + {1 \over 3}x^{-{1 \over 6}}\cos(x^{1 \over 3})$. The limit as $x$ goes to infinity of $f(x)$ doesn't exist; the $x^{1 \over 2}$ factor increases to infinity while the sine factor modulates it. On the other hand, the limit of $f'(x)$ is zero.

edited Jun 17 '11 at 16:01

answered Jun 17 '11 at 15:40

Zarrax

44,950

what you defined doesn't have a limit at infinity, both $f$ and $f'$. – Jun 17 '11 at 15:53
ok corrected it, this should work – Zarrax Jun 17 '11 at 15:55
4

You don't need the $\sqrt{x}$. Just take $f(x) = \sin(x^p)$ where $0 < p < 1$. – Robert Israel Jun 17 '11 at 19:33

score 0 · Answer 2 · answered Jul 18 '20 at 15:03

Take the example $f(x)=\frac{1}{x}+\sin(\sqrt x)$ ,$x\in (1,\infty)$

So,$f'(x)=-\frac{1}{x^2}+\frac{\cos\sqrt x}{2\sqrt x}$ ,$x\in (1,\infty)$

Thus,$\lim\limits_{x\to\infty} f'(x)=0<\infty$ but $\lim\limits_{x\to \infty} f(x)$ does not exist.

So,we are done.

If $f$ is differentiable in $(1,\infty )$ and $\lim_{x\to\infty }f'(x)=L<\infty $ then $\lim_{x\to\infty }f(x)=l\le\infty$?

2 Answers2