Assume that $f:{\bf R}\to{\bf R}$ is differentiable on ${\bf R}$, and both of $\lim\limits_{x\to\infty}f(x)$ and \lim\limits_{x\to\infty}f'(x) are finite. Geometrically, one may have \lim_{x\to\infty}f'(x)=0 Here is my question:
How can one actually prove it?
By definition, it suffices to show that
$\lim_{x\to\infty}\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=0$
i.e. $\forall \epsilon>0~\exists M>0\quad \textrm{s.t.}\quad x>M\Rightarrow \left|\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\right|<\epsilon$ For large enough $M$ and small enough $\epsilon$, one has $|f(x+h)-f(x)|<\tilde{\epsilon}$ But I have no idea how to go on.