A calculus question On the interval $(0, \infty)$,the function $f \geq 0$,$f' \leq 0$, and $f'' \geq 0$.Prove that $\lim\limits_{x \to \infty} xf'(x) = 0$. 
 A: If the limit is not zero, there is an $\epsilon>0$ such that $xf'(x) \le -\epsilon$ at arbitrarily large $x$. So we can construct a sequence $x_n$ such that $x_n\ge2^n$, $x_{n+1}\ge 2x_n$ and $x_nf'(x_n) \le -\epsilon$ for all $n$. But then the lower halves of the rectangles that the points $(x_n,-\epsilon/x_n)$ form with the origin all lie above the graph of $f'$ (i.e. between it and the $x$ axis), are all disjoint, and all have area $\epsilon/2$. It follows that the indefinite integral of $f'$, i.e. $f$, diverges to $-\infty$, in contradiction with $f\ge0$.
A: 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.
A: For $x>1$, $f(x)$ is bounded below by $0$ and above by $f(1)$.  Therefore $\lim_{x\to\infty} {f(x)\over \log x}=0$.  By l'Hopital's rule, $\lim_{x\to\infty} xf'(x) =0$ as well.
