TITLE: If a function has a finite limit at infinity, does that imply its derivative goes to zero?
QUESTION [40 upvotes]: I've been thinking about this problem: Let $f: (a, +\infty) \to \mathbb{R}$ be a differentiable function such that $\lim\limits_{x \to +\infty} f(x) = L < \infty$. Then must it be the case that $\lim\limits_{x\to +\infty}f'(x) = 0$?
It looks like it's true, but I haven't managed to work out a proof. I came up with this, but it's pretty sketchy:
$$
\begin{align}
\lim_{x \to +\infty} f'(x) &= \lim_{x \to +\infty} \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \\
&= \lim_{h \to 0} \lim_{x \to +\infty} \frac{f(x+h)-f(x)}{h} \\
&= \lim_{h \to 0} \frac1{h} \lim_{x \to +\infty}[f(x+h)-f(x)] \\
&= \lim_{h \to 0} \frac1{h}(L-L) \\
&= \lim_{h \to 0} \frac{0}{h} \\
&= 0
\end{align}
$$
In particular, I don't think I can swap the order of the limits just like that. Is this correct, and if it isn't, how can we prove the statement? I know there is a similar question already, but I think this is different in two aspects. First, that question assumes that $\lim\limits_{x \to +\infty}f'(x)$ exists, which I don't. Second, I also wanted to know if interchanging limits is a valid operation in this case.

REPLY [12 votes]: Because all counterxamples given here are oscillating, perhaps someone might wonder whether the proposition is true if we require the funcion to be monotonic. The answer is still no. Simply consider the inverse of the function, monotonic on $(0,\infty)$, given here $f(x)=\frac{1}{x}+\sin\left(\frac{1}{x}\right)$.
Also: consider the cumulative distribution function corresponding to 
a probability density function that has no limit as $x\to +\infty$. For example, the integral of the density constructed here. Or, using the same idea with a Cauchy distribution we get other example:
$$F(x)= \frac{2}{\pi}\sum_{k=1}^\infty \frac{\tan^{-1}\left((x-k) \pi 2^k\right)}{2^k} $$

This tends to $1$, but its derivative $f(x)=F'(x)= 2 \sum_{k=1}^\infty (\pi^2 2^{2k} (x-k)^2 +1)^{-1} $ has no limit, because $f(n)>2$ $\forall n \in \mathbb{N}$.