TITLE: Prove $\lim \limits_{x \to\infty } \frac{f(x)}{x}=0$ and $f$ differentiable implies $ \lim \limits_{x \to\infty } \inf |f'(x)|=0 $
QUESTION [8 upvotes]: Given a differentiable function on $(a,+\infty)$ such as $\lim \limits_{x \to\infty } \frac{f(x)}{x}=0$ prove the following:
$$ \lim \limits_{x \to\infty } \inf |f'(x)|=0  $$
I just can't see how to do it...
(even after understanding How to show that $\lim\limits_{x \to \infty} f'(x) = 0$ implies $\lim\limits_{x \to \infty} \frac{f(x)}{x} = 0$?)

REPLY [7 votes]: It suffices to show that there exists a sequence $x_n\to \infty$ such that $f'(x_n)\to 0$. 
By the mean value theorem, for each $n$ there exists $x_n$ with $n<x_n<2n$ such that 
$$
f'(x_n) = \frac{f(2n)-f(n)}{2n-n}=\frac{f(2n)-f(n)}{n}=2\frac{f(2n)}{2n}-\frac{f(n)}{n}.
$$
Since $\lim_{x\to \infty} f(x)/x=0$, it follows that the right hand side tends to zero, and hence
$$
\lim_{n\to \infty} f'(x_n)=0.
$$
By the squeeze theorem, the fact that $x_n>n$ implies that 
$$
\lim_{n\to \infty} x_n=\infty.
$$
This finishes the proof.