Limit of $f'(x) e^{-f(x)}$ Let $f$ be a real function verifying $f''\geq C>0$, where C is a constant.
 Do we have : $\lim_{x\to +\infty}f'(x) e^{-f(x)}=0$ ?

Hint: transform the limit to an indeterminate form such as $[\frac{0}{0}]$ or $[\frac{\infty}{\infty}]$, and apply De L'Hopital's Rule:
as you mentioned in a comment $\lim_{x\to +\infty}f'(x)=+\infty$ and $\lim_{x\to +\infty}f(x)=+\infty$, so $\lim_{x\to +\infty}f'(x) e^{-f(x)}$ would be an indeterminate form such as $+\infty\cdot0$. You'd better to turn the limit in order to use De L'Hopital's Theorem:
$$\lim_{x\to +\infty}f'(x) e^{-f(x)}=\lim_{x\to +\infty}\frac{e^{-f(x)}}{\frac{1}{f'(x)}}=\lim_{x\to +\infty}\frac{-f'(x)e^{-f(x)}}{\frac{-f''(x)}{f'(x)^2}}=...$$
