Existence of limit at infinity There is a given real function which is differentiable and such that 
$$\lim_{x\to\infty}f'(x)=L$$
We're asked to give a proof that if $\;f\;$ is bounded, then the limit must be $\;L=0\;$
What I worked out: geometrically it looks very clear, but I'm having problems with the formal part.
I said: we suppose $\;|f(x)|\le M\;\;\;\forall\ x\in\Bbb R\;$ , and then I tried the mean value theorem. Since
$$\lim_{x\to\infty}\frac{f(x)-f(0)}x=0\;,\;\;\text{because}\;\;|f(x)-f(0)|\le2M\;\;\text{(bounded)}$$
there exists $\;c_x\in(o,x)\;$ for all positive $\;x\;$, so that
$$\frac{f(x)-f(0)}x=f'(c_x)$$
The problem I meet is that if I do$\;x\to\infty\;$ I cannot deduce that $\;c_x\to \infty\;$ because I don't know if we could be getting the same $\;c_x\;$ for different $\;x$'s. If I could solve this then I would get immediately that $\;L=0\;$ .
Any help will be appreciated.

Sketch of proof.  Suppose that $L\ne0$, and without loss of generality that $L>0$.  Then there exists $x_0$ such that if $x>x_0$, then $f'(x)>L/2$.  Using the mean value theorem, if $x>x_0$ then
$$f(x)=f(x_0)+f'(c)(x-x_0)>f(x_0)+\frac L2(x-x_0)\ ,$$
and the RHS is unbounded as $x\to\infty$.
See if you can fill in the details.
