limit of $f$ and $f''$ exists implies limit of $f'$ is 0 
Prove that if $\lim\limits_{x\to\infty}f(x)$ and $\lim\limits_{x\to\infty}f''(x)$ exist, then $\lim\limits_{x\to\infty}f'(x)=0$.

I can prove that $\lim\limits_{x\to\infty}f''(x)=0$. Otherwise $f'(x)$ goes to infinity and $f(x)$ goes to infinity, contradicting the fact that $\lim\limits_{x\to\infty}f(x)$ exists. I can also prove that if $\lim\limits_{x\to\infty}f'(x)$ exists, it must be 0. So it remains to prove that $\lim\limits_{x\to\infty}f'(x)$ exists. I'm stuck at this point.

Hint $\ $ This follows easily from L'Hôpital's rule since
$$\rm \lim_{x\to\infty}\ f-f'\ =\ \lim_{x\to\infty}\frac{e^x\ (f-f')}{e^x}\ =\ \lim_{x\to\infty}\frac{e^x\ (\:f-f'+f'-f'')}{e^x}\ =\ \lim_{x\to\infty}\ f-f''\ exists$$
See also the similar classic Hardy old problem.
