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?

Here's a proof that avoids L'Hôspital's rule (though you can certainly use it, as noted by Bill; basically, the condition on $f'(x)$ ensures that the function cannot oscillate like crazy).
Let $\epsilon\gt 0$. We want to show that there exists $N\gt 0$ such that if $x\gt N$, then
$$\left|\frac{f(x)}{x}\right| \lt\epsilon.$$
We know that there exists $M\gt 0$ such that $|f'(x)|\lt \frac{\epsilon}{2}$ for all $x\gt M$. By the Mean Value Theorem, if $x\gt M$, there exists $c\in (M,x)$ such that $f(x)-f(M)=f'(c)(x-M)$, so
$$|f(x)-f(M)| = |f'(c)|(x-M) \lt \left(\frac{\epsilon}{2}\right) x.$$
That is, for all $x\gt M$,
$$\left|\frac{f(x)}{x} - \frac{f(M)}{x}\right| \lt\frac{\epsilon}{2}.$$
As $x\to\infty$, we know $\frac{f(M)}{x}\to 0$ (since $f(M)$ is fixed). So there exists $N\gt M$ such that if $x\gt N$, then $\left|\frac{f(M)}{x}\right|\lt \frac{\epsilon}{2}$. Thus, if $x\gt N$, then
$$\left|\frac{f(x)}{x}\right| \leq \left|\frac{f(x)-f(M)}{x}\right| + \left|\frac{f(M)}{x}\right| \lt \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon,$$
which is what we needed to prove. QED
