We want infinitely differentiable monotonic function f such that: $\lim_{x \to +\infty} f(x) = 0$ and $\lim_{x \to +\infty} f'(x) \neq 0$.
An infinitely differentiable bridging function, strictly monotonic on $[0,1]$ which will help us in our example:
$$f(x) = \begin{cases} 0 & x \leq 0 \\
\exp(-\frac{1}{x^2}\exp(\frac{-1}{(1-x)^2})) & 0 < x < 1 \\
1 & x\geq1
\end{cases}$$
Let f(x) be equal 1 for $x\leq1$, equal $\frac{1}{n}$ on the closed interval $[2n-1,2n]$ for $n=1,2,…$ and on the intervening intervals $(2n,2n+1)$ define f(x) by translations of the bridging functions with appropriate negative factors for changes in the vertical scale.
My question is why $\lim_{x \to +\infty} f'(x) \neq 0$? On intervals $[2n-1,2n] f'(x)=0$, and between we have this bridge that connects two such intervals and I see that derivative is not 0 there but it decreases when $x$ increases since the bridge has to "travel" $\frac{1}{n(n+1)}$ in vertical and 1 distance in horizontal, so I think $f'(x)$ goes to 0 at infinity.
It is an example from the book counterexamples in analysis.
Best Answer
I agree with you, this doesn't seem to work. But if you scale the non-constant intervals in the $x$-direction by a factor of $\alpha$, then you increase $f'(x)$ by a factor of $1/\alpha$. So set $\alpha=\frac{1}{n(n+1)}$ to have $f'(x)$ not tend to zero; set $\alpha=\frac{1}{n^3}$ to have $f'(x)$ be unbounded. I expect this is what they had in mind.