Let $f'\ge0, f''\ge0 \ \forall x \in \mathbf{R}$. If there is $c$ such that $f'(c)>0$, then $\lim_{x\to\infty}f(x)=\infty$ 
Let $f$ be a twice differentiable function. Suppose $f'(x)\ge0, f''(x)\ge0 \ \forall x \in \mathbf{R}$. If there exists $c \ \in R$ such that $f'(c)>0$, show that $\lim_{x\to\infty}f(x)=\infty$. My attempt is to prove the statement by absurd. So, I would like to have a feedback on my proof, please. Setting myself that $f(c_x)>0$ in my proof seems to me a little bit strange..

Suppose by absurd that $f$ converges to $L$: $\lim_{x\to\infty}f(x)=L$. As $f''\ge0$, then $f$ is convex and we can deduce that $f'$ is increasing.
Consider the interval $[x,x+a]$ with $a>0$.  As $f$ is differentiable on $\mathbf{R}$, it is differentiable on $[x,x+a]$ and so continuous on $[x,x+a]$. By mean value theorem there exists $c_x\in ]x,x+a[$  such that: $f'(c_x)=\frac{f(x+a)-f(x)}{a}$. As $x$ is "arbitrary" in the interval, we can suppose that $f'(c_x)=\frac{f(x+a)-f(x)}{a}>0$. But,
$f'(c_x)=\frac{f(x+a)-f(x)}{a} \iff \underbrace{\lim_{x\to\infty}f'(c_x)}_{(*)}=\lim_{x\to\infty}\frac{f(x+a)-f(x)}{a}=\frac{L-L}{a}=0\ngtr 0$. So we got a contradiction. Thus, $f$ diverges.
(*) As $f'$ is increasing, as $x$ goest to infinity the inequality might hold and so $\lim_{x\to\infty}f'(c_x)>0$ might hold.

I think your proof is fine, but the idea behind the proof is somewhat obscured. To me, it seems the key point is to prove there is a line of positive slope that bounds $f$ from below. Indeed, if $f'(c) > 0$, then $f(x) - f(c) = f'(c_x)(x-c) \geq f'(c)(x-c)$. The right side clearly tends to $\infty$ as $x \to \infty$, hence so must $f$.
Edit, I should also point out that this proof also works for convex $f$ (not assumed to be twice differentiable), as convex implies $f$ is above any tangent line, which is all we really used.
