TITLE: Proving the limit at $\infty$ of the derivative $f'$ is $0$ if it and the limit of the function $f$ exist.
QUESTION [5 upvotes]: Suppose that $f$ is differentiable for all $x$, and that $\lim_{x\to \infty} f(x)$ exists.
Prove that if $\lim_{x\to \infty} f′(x)$ exists, then $\lim_{x\to \infty} f′(x) = 0$, and also, give an example where $\lim_{x\to \infty} f′(x)$ does not exist.
I'm at a loss as to how to prove the first part, but for the second part, would a function such as $\sin(x)$ satisfy the problem?

REPLY [3 votes]: Assuming both limits exist, apply L'Hospitals rule: 
$$L= \lim_{x \rightarrow \infty}f(x) = \lim_{x \rightarrow \infty}\frac{e^xf(x)}{e^x}\\=\lim_{x \rightarrow \infty}[f(x) + f'(x)] = L + \lim_{x \rightarrow \infty} f'(x) , $$
and conclude that $ \lim_{x \rightarrow \infty} f'(x)= 0$.
Alternatively, by the MVT there is a point $\xi_x \in(x,x+1)$ such that
$$ f(x+1)-f(x)=f'(\xi_x)$$
If $f'(x) \rightarrow L'$, then for every $\epsilon > 0$ there is a $K>0$ such that $|f'(x) - L'| < \epsilon$ when $x > K$. As $\xi_x > x > K$, it follows that $|f'(\xi_x) - L'| < \epsilon.$
Hence,
$$ L'=\lim_{x \rightarrow \infty}f'(x)=\lim_{x \rightarrow \infty}f'(\xi_x)=\lim_{x \rightarrow \infty}[f(x+1)-f(x)]=0.$$