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.
Best Answer
The answer is: No. Consider $f(x)=x^{-1}\sin(x^3)$ on $x\gt0$. The derivative $f'(x)$ oscillates between roughly $+3x$ and $-3x$ hence $\liminf\limits_{x\to+\infty}\,f'(x)=-\infty$ and $\limsup\limits_{x\to+\infty}\,f'(x)=+\infty$.