# 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 $$x_n\rightarrow +\infty.$$

The Mean Value Theorem says that for any $$x$$, there is a $$\xi\in(x,2x)$$ so that $$\frac{f(2x)-f(x)}{x}=f'(\xi)\tag1$$ Choose any $$\epsilon\gt0$$. Find an $$M$$ so that $$x\ge M\implies\left|\frac{f(x)}{x}\right|\lt\frac{\epsilon}3$$. Then for $$x\ge M$$, \begin{align} \left|f'(\xi)\right| &=\left|\,\frac{f(2x)-f(x)}{x}\,\right|\\ &=\left|\,2\frac{f(2x)}{2x}-\frac{f(x)}x\,\right|\\ &\le2\left|\,\frac{f(2x)}{2x}\,\right|+\left|\,\frac{f(x)}x\,\right|\\[6pt] &\le\epsilon\tag2 \end{align} Since $$\epsilon\gt0$$ was arbitrary, we can find a sequence $$M_n\ge n$$ which corresponds to $$\epsilon=2^{-n}$$, which gives a sequence $$\xi_n\in(M_n,2M_n)$$ so that $$|f'(\xi_n)|\le2^{-n}$$.
Suppose it is not true, there exists $$c>0$$ and $$M$$ such that $$x>M$$ implies that $$|f'(x)|>c$$, if $$x>M, f(x)-f(M)=f'(y)(x-M)$$ implies that $$|{{f(x)}\over x}|$$ $$\geq c-|{{f(M)}\over x}|$$ contradiction with the fact that $$lim_{x\rightarrow+\infty}|{{f(x)}\over x}=0$$.
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 $$n \to \infty$$].