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.
 A: $f'(x)>0$ tells you that $f$ is increasing and $f''(x)>0$ tells you that $f'$ is also increasing. I will do the case for $+\infty$, you can do the other one by yourself.
So pick $x\in\mathbb{R}$. From Lagrange´s theorem, you know that for each $y\geq x$
$f(y)-f(x)=f'(c)(y-x)$ for some $c\in(x,y)$.
But $f'$ is non decreasing, so $0<f'(x)\leq f'(c)$ (actually it is greater, not equal since $f''>0$). Combining these two, we can write 
$f(y) - f(x)\geq f'(x) (y-x) .$
From this, it is clear that $\lim_{y\to +\infty} f(y)-f(x) = +\infty$, so
$\lim_{y\to +\infty} f(y)=+\infty$.
A: Hint: suppose you let 
$$
g(x) = \frac{f(x)}{C}
$$
where $C$ is positive, and you show that $\lim_{x\to \infty} g(x) = \infty$. Then you'd be done, right? 
Try this with $C = f'(0)$ (which you know to be positive). What can you say about $g'(t)$ for $t > 0$? What's that tell you about $g(t)$, for $t > 0$. 
Hint: Mean value theorem. (Unless you already know about integration, in which case: Hint: integrate.)
A: Suppose that $f'(0)=a>0$. Since $f''(x)>0$, then $f'(x)>a$ for all $x>0$ (otherwise by MVT there is a point where $f''(x)\le 0$). Then $f(x)\ge ax+f(0)$ for all $x\ge 0$ - to argue this just consider $g(x)=f(x)-ax-f(0)$ and draw out a contradiction assuming $g<0$ for some $x$. Then since it's plainly true that $ax+f(0)\to \infty$ ($a>0$) you have the result
A: Note that $f''$ is positive and hence $f'$ is strictly increasing and hence $f'(x) $ tends to a limit $L$ or to infinity. Moreover if it tends to $L$ then $L\geq f'(0)>0$. Again using the same logic we can see that either $f(x) $ tends to $M$ or to infinity. If $f(x) \to M$ then $f(x+1)-f(x)=f'(\xi)$ tends to $0$ which contradicts that $f'$ tends to $L>0$ or to infinity. It follows that $f(x) \to\infty$ as $x\to\infty$. 
A: 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 $f(y)\to_{y\to\infty}\infty$
