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

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.

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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.
what you defined doesn't have a limit at infinity, both $f$ and $f'$. – timhortons Jun 17 '11 at 15:53
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