How to show that $\lim\limits_{x \to \infty} f'(x) = 0$ implies $\lim\limits_{x \to \infty} \frac{f(x)}{x} = 0$? I was trying to work out a problem I found online. Here is the problem statement:

Let $f(x)$ be continuously differentiable on $(0, \infty)$ and suppose $\lim\limits_{x \to \infty} f'(x) = 0$. Prove that $\lim\limits_{x \to \infty} \frac{f(x)}{x} = 0$. 

(source: http://www.math.vt.edu/people/plinnell/Vtregional/E79/index.html)
The first idea that came to my mind was to show that for all $\epsilon > 0$, we have $|f(x)| < \epsilon|x|$ for sufficiently large $x$. (And I believe I could do this using the fact that $f'(x) \to 0$ as $x \to \infty$.)
However, I was wondering if there was a different (and nicer or cleverer) way. Here's an idea I had in mind:
If $f$ is bounded, then $\frac{f(x)}{x}$ clearly goes to zero. If $\lim\limits_{x \to \infty} f(x)$ is either $+\infty$ or $-\infty$, then we can apply l'Hôpital's rule (to get $\lim\limits_{x \to \infty} \frac{f(x)}{x} = \lim\limits_{x \to \infty} \frac{f'(x)}{1} = 0$). 
However, I'm not sure what I could do in the remaining case (when $f$ is unbounded but oscillates like crazy). Is there a way to finish the proof from here? 
Also, are there other ways of proving the given statement?

You can do a proof by contradiction. If ${\displaystyle \lim_{x \rightarrow \infty} {f(x) \over x}}$ were not zero, then there is an $\epsilon > 0$ such that there are $x_1 < x_2 < ... \rightarrow \infty$ such that ${\displaystyle\bigg|{f(x_n) \over x_n}\bigg| > \epsilon}$. Then for $a \leq x_n$ one has
$$|f(x_n) - f(a)| \geq |f(x_n)| - |f(a)|$$
$$\geq \epsilon x_n - f(a)$$
By the mean value theorem, there is a $y_n$ between $a$ and $x_n$ such that
$$|f'(y_n)| = {|f(x_n) - f(a)| \over x_n - a}$$
$$\geq {\epsilon x_n - f(a) \over x_n - a}$$
Letting $n$ go to infinity this means for sufficiently large $n$ you have
$$|f'(y_n)| > {\epsilon \over 2}$$
Since each $y_n \geq a$ and $a$ is arbitrary, $f'(y)$ can't go to zero as $y$ goes to infinity.
