Twice differentiable function to infinity Let $f: \mathbb R \to \mathbb R$ be twice differentiable function, to which both $f'(x) > 0$ and $f''(x) > 0$ for all $x \in \mathbb R$. Show that $\lim_{x\to\infty}$ $f(x) = \infty$.
Tried using the definitions of differentiation but got nowhere.

We can even show that $f$ grows faster than linearily.
By the Mean value theorem, for all $x<y\in\mathbb R$ there is $u\in [x,y]$ such that $$f(y) = f(x) + f'(u)(y-x).$$ Of course, this applies to $f'$ as well, so $$f'(u) = f'(x) + f''(t)(u-x)$$ for some $t\in[x,u]$.
Plugging in $x=0$:
$$f(y) = f(0) + f'(u)\cdot y = f(0) + \left(f'(0) + \underbrace{f''(t) \cdot u}_{>0}\right)\cdot y > f(0) + f'(0)\cdot y\text{ for all }y>0$$
Since $f'(0) > 0$, this yields $f(y)\to_{y\to\infty}\infty$
