TITLE: A calculus question
QUESTION [6 upvotes]: On the interval $(0, \infty)$,the function $f \geq 0$,$f' \leq 0$, and $f'' \geq 0$.Prove that $\lim\limits_{x \to \infty} xf'(x) = 0$.

REPLY [2 votes]: On the one hand, $f$ is monotone decreasing (since $f' \leq 0$) and $f \geq 0$; hence, for some $l \geq 0$,
$\lim _{x \to \infty } f(x) = l$.
On the other hand, $f'$ is monotone increasing (since $f'' \geq 0$) and $f' \leq 0$; hence 
$$
f(x) - f(x/2) = \int_{x/2}^x {f'(u)du}  \le \int_{x/2}^x {f'(x)du}  = \frac{{x f'(x)}}{2} \le 0.
$$
Letting $x \to \infty$, the left-hand side converges to $0$; hence $x f'(x) \to 0$ too.