Does the derivative of a bounded smooth monotone function have a limit at infinity?

Question

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?

No. Let $f$ increase to $1$, but approximately constant most of the time except for "abrupt" (but smooth) jumps. — David Mitra, Mar 15 '14 at 19:50
What you need to require on $f$ is that the limit of $f'$ exists. — Git Gud, Mar 15 '14 at 19:52
@DavidMitra I guess you're function is a counter example of the case in which we do not require the existance of $\lim f'$, right? — Ant, Mar 15 '14 at 19:56
@Ant Here is a proof that will make you want to cry from how beautiful it is. — Git Gud, Mar 15 '14 at 19:59
Assuming the limit exists, apply the Mean Value Theorem to $f$ on the intervals $[n,n+1]$. — David Mitra, Mar 15 '14 at 19:59
possible duplicate of Limit of the derivative of a function as x goes to infinity — Caleb Stanford, Mar 15 '14 at 20:05
I disagree that it is a duplicate. The hypothesis the OP assumes aren't the same and require different counterexamples than the usual ones. — Git Gud, Mar 15 '14 at 20:06
@DavidMitra if I apply the mean value theorem we have $f(n+1) - f(n) = f'(c)$ , $c \in [n, n+1]$. Taking the limit (if $\lim f'$ exists) we have $\lim f' = 0$. So the hypothesis of monotonicity can be relaxed.. Do we still need to require that $f'(x)$ is continuos ? — Ant, Mar 15 '14 at 20:11

score 4 · Accepted Answer · edited Apr 13 '17 at 12:20

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.

thank you! Very nice, I figured what was wrong but having an actual example is great :) — Ant, Jul 11 '14 at 07:32

Does the derivative of a bounded smooth monotone function have a limit at infinity?

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