If derivative goes to $0$, does the function have a limit?

Question

Let $f:\mathbb{R}\to\mathbb{R}$ be a differentiable function. Assume that $\lim_{x\to\infty}f'(x)=0$, does that mean that $\lim_{x\to\infty}f(x)$ exists?

PS. What if we assume that $f$ is also bounded?

Answer 1

Not necessarily. Consider $f(x)=\sin(\ln(x))$. Then $f'(x)=\frac1x\cos(\ln(x))\to0$, but $f(x)$ diverges, as it reaches $-1$ and $1$ infinitely often.

Answer 2

Not necessarily. See $f(x)=\ln x$:

$$\lim_{x\to \infty}f'(x)=\lim_{x\to\infty}\frac 1 x=0$$ $$\lim_{x\to\infty}f(x)=+\infty$$

Answer 3

No. Consider the function defined by cos(x) on [0,2pi], cos((x/2)-pi) on [2pi,6pi] and so on.

Answer 4

No.

Consider $\lim_{t\to\infty}f'(t)=0$

and

$\lim_{x\to\infty}f(x)-f(a)=\lim_{x\to\infty}\int_{a}^{x}f'(t)dt$ and $a\geq0$.

In this case, that integral converges if $f(a)$ exists for all $a\geq0$. for example the function $f(x)=\ln x$ doesn't satisfy that condition.

And if that integral converges, the limit exists.