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?

let $f:\left[ a,+\infty \right[ \longrightarrow \mathbb{R}  $ differentiable suppose that  $\displaystyle \lim_{x\rightarrow +\infty} f'(x)=l \in \mathbb{R}$ prove that $\displaystyle \lim_{x\rightarrow +\infty} \dfrac{f(x)}{x}=l $ 
we will begin with the mean value theorem : for $ x > \max(0,a)  \ \exists c\in \left]a,x\right[ $ such that
    $$f(x)-f(a)=f'(c)(x-a)\implies \dfrac{\dfrac{f(x)}{x}-\dfrac{f(a)}{x}}{1-\dfrac{a}{x}}=f'(c)$$
    and because as $c\rightarrow +\infty$ , $x\rightarrow +\infty$ then we have:
    $$\lim_{c\rightarrow +\infty} f'(c)=\lim_{c\rightarrow +\infty }\dfrac{\dfrac{f(x)}{x}-\dfrac{f(a)}{x}}{1-\dfrac{a}{x}}\implies \lim_{c\rightarrow +\infty}\dfrac{\dfrac{f(x)}{x}-\dfrac{f(a)}{x}}{1-\dfrac{a}{x}}=l\implies \lim_{x\rightarrow +\infty}\dfrac{\dfrac{f(x)}{x}-\dfrac{f(a)}{x}}{1-\dfrac{a}{x}}=l$$
thus we have the desired result:
    $$\lim_{x\rightarrow +\infty} \dfrac{f(x)}{x}=l$$
