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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?

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    $\begingroup$ By L'Hospital's Rule we get $f(x) /x\to 0$ and that's the best we can conclude here. $\endgroup$ – Paramanand Singh Dec 29 '18 at 1:37
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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.

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

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

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  • $\begingroup$ ah ok my bad. I missed this example. I wonder what if $f$ is also bounded. $\endgroup$ – David Lingard Dec 28 '18 at 23:00
  • $\begingroup$ Maybe I should edit the question and add $f$ bounded. $\endgroup$ – David Lingard Dec 28 '18 at 23:01
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No. Consider the function defined by cos(x) on [0,2pi], cos((x/2)-pi) on [2pi,6pi] and so on.

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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.

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