TITLE: Does the derivative of a bounded smooth monotone function have a limit at infinity?
QUESTION [5 upvotes]: Let $f \in C^1(\mathbb{R})$ a monotonic function such that $$\lim_{x \to \infty} f(x) = m \in \mathbb{R}$$
Does this imply $\displaystyle\lim_{x \to \infty} f'(x) = 0$? 
If so, can the hypothesis be relaxed?
If not, which additional condition we must require on $f(x)$ so that it works?
EDIT:
Let's require also that $\displaystyle\lim_{x \to \infty} f'(x)$ exists.
How can one prove the claim?

REPLY [4 votes]: To construct a counterexample, it is easier to begin with the derivative of $f$. Define 
$$
g(x) = \sum_{n=1}^\infty (1-2^n|x-n|)^+
$$
where $a^+=\max(a,0)$. The graph of $g$ consists of triangular spikes at every positive integer; the $n$th spike has height $1$ and base $2^{1-n}$, hence its area is $2^{-n}$. Define 
$$
f(x) = \int_0^x g(t)\,dt
$$
Since $g$ is continuous and nonnegative, $f$ is $C^1$ and monotone. Also, for every $x$
$$f(x)\le \sum_n 2^{-n}=1$$
which implies that $\lim_{x\to\infty} f(x)$ exists (in fact, it is equal to $1$). 
On the other hand, $f'=g$, which does not have a limit at infinity.
The proof of $\lim f'=0$ under the additional assumption that $\lim f'$ exists was already given.