If $\lim_{x\rightarrow +\infty} \frac{f(x)}{x}=0,$ show there is $x_n\rightarrow\infty$ such that $\lim_{n\rightarrow\infty}f'(x_n)=0.$ I'm working on the problem:
Suppose $f(x)$ is differentiable on $(0,+\infty)$.
If $$\lim_{x\rightarrow +\infty} \frac{f(x)}{x}=0,$$ show there is 
$x_n\rightarrow\infty$ such that $$\lim_{n\rightarrow\infty}f'(x_n)=0.$$
Here are some of my thoughts: 
Let $n\geq 2$, then $\frac{f(n)-f(1)}{n-1}=f'(x_n)$ for some $x_n\in (1,n)$,  so $f'(x_n)\rightarrow 0$. But I can't show $x_n\rightarrow +\infty.$

If the conclusion is not true then there exists $\epsilon >0$ and $M$ such that $|f'(x)| >\epsilon$ for  all $x \geq M$. Using the fact that derivatives have IVP we see that we can actually make  $f'(x) >\epsilon$ for  all $x \geq M$ or $f'(x)<-\epsilon$ for  all $x \geq M$. Consider the former case. Note that $f(n+1)-f(n) \geq \epsilon $ for all $n >M$ by MVT. Now you can see easily that $\frac  {f(n)} n $ does not tend to $0$.
[If $n_0 >M$ then $f(n) \geq (n-n_0)\epsilon+f(n_0)$ for all $n >n_0$ so $\frac {f(n)} n \geq (1-\frac {n_0} n)\epsilon+\frac 1 nf(n_0) \to \epsilon$ as $n \to \infty$].
