If $\lim\limits_{x \to \pm \infty} \frac{f(x)}{x} = +\infty$ then $\lim\limits_{x \to \pm \infty} \frac{\int_0^x f(s) \; ds}{x} = +\infty$ 
Let $f \in C(\mathbb{R})$
If $\lim\limits_{x \to \pm \infty} \frac{f(x)}{x} = +\infty$ and $F(x) = \int_0^x f(s) \; ds$ then $\lim\limits_{x \to \pm \infty} \frac{F(x)}{x} = +\infty$

I thought of using mean value theorem but this would give me a $c \in ]0,x[$ and I need it to diverge. 
Edit
I also tried l'Hôpital but for that I should have $F(x)$ diverging and this needs not be the case.
What other idea could I use here?

$$ \lim_{x \to \pm \infty} \frac{f(x)}{x} = \infty \Rightarrow \lim_{x \to \pm \infty} f(x) = \pm \infty.$$
Now use L'Hopital's Rule on your problem:
$$ \lim_{x \to \pm \infty} \frac {\int_0^x f(s)~ds}{x}= \lim_{x \to \pm \infty} \frac{f(x)}{1}= \pm \infty.$$
