# Prove $\lim \limits_{x \to\infty } \frac{f(x)}{x}=0$ and $f$ differentiable implies $\lim \limits_{x \to\infty } \inf |f'(x)|=0$

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 $\lim\limits_{x \to \infty} \frac{f(x)}{x} = 0$?)

• For some $\epsilon > 0,$ what happens if $f'(x) > \epsilon$ for all $x > X_0,$ where $X_0$ is some large positive number? The other case is, what happens if $f'(x) < - \epsilon$ with everything else written the same? Feb 25, 2013 at 5:40
• @WillJagy than f(x) is monotonically increasing to infinity, but it doesn't yet a contradict
– User
Feb 25, 2013 at 5:46
• Anyway, there is still a little more to it after that, but it is a start. Feb 25, 2013 at 5:46
• User, yes, monotone to infinity, and what about $f(x) / x?$ Feb 25, 2013 at 5:47
• Or, what about $(f(x)-f(X_0)) / (x - X_0)?$ Feb 25, 2013 at 5:54

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.
In case $f'(x)>\epsilon$ for all $x\geq x_0$ then (show that) $f(x)>f(x_0)+\epsilon(x-x_0)$ for all $x\geq x_0$.
Then show that $f(x)>f(x_0)+\epsilon(x-x_0)>\frac{\epsilon}{2}x$ for all $x\geq x_1$ for some $x_1\geq x_0$...