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Clarification: Does $\lim\limits_{x \to \infty}f'(x)=0$ as $x$ approaches infinity mean $\lim\limits_{x \to \infty} f(x)$ exists in the extended real numbers $[-\infty,\infty]$?

I was thinking a lot about this problem and I couldn't prove it; I would appreciate any help!

This isn't a duplicate! In that problem you have to prove that the limit of f(x) exists in the extended real numbers, you don't assume it does.

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marked as duplicate by alexjo, JonMark Perry, Leucippus, Xander Henderson, Jack D'Aurizio calculus Oct 8 '17 at 2:07

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ You can get any behavior at infinity combined with any behavior at zero. If you really do mean limits at $\infty$ and at $0$. $\endgroup$ – GEdgar Oct 7 '17 at 20:59
  • $\begingroup$ No, the thing is that in that problem you need to prove that the limit of f(x) exists (including infinity) $\endgroup$ – Tamir Oct 7 '17 at 21:02
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    $\begingroup$ @Tamir The limit of $f(x)$ as $x$ goes to what? did you mean to ask about $\lim_{x\to\infty}$ or $\lim_{x\to0}$? $\endgroup$ – John Doe Oct 7 '17 at 21:03
  • $\begingroup$ as x approaches infinity $\endgroup$ – Tamir Oct 7 '17 at 21:05
  • $\begingroup$ Does anyone have any idea? $\endgroup$ – Tamir Oct 8 '17 at 9:14
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No, there is a simple counter example. $$f'(x)=\frac1x\implies \lim_{x\to\infty}f'(x)=0$$ But $$f(x)=\log x, \lim_{x\to\infty}f(x)=\infty$$

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    $\begingroup$ he asked about $\lim \limits_{x \to 0} f(x)$, furthermore he asks what is the general behavior. $\endgroup$ – Ahmad Oct 7 '17 at 20:57
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    $\begingroup$ @Ahmad Hmm, I am sure he must have meant $\infty$, especially after reading the title. If he did mean $x\to0$, then you could take $f(x)=\frac1x$ to get infinite behaviour at $0$, and something like $f(x)=e^{-x}$ to get finite behaviour. But as mentioned in the comment on the question, you can combine functions to get any behaviour at $0$, regardless of what happens at $\infty$ $\endgroup$ – John Doe Oct 7 '17 at 21:01
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    $\begingroup$ @Tamir I answered your question with $x\to\infty$. My comment referred to $x\to 0$. $\endgroup$ – John Doe Oct 7 '17 at 21:07
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    $\begingroup$ @Tamir I gave you an example of a function $f(x)$ whose derivative goes to $0$ as $x\to\infty$, but where the actual function $f(x)$ does not converge to a finite value - it goes to infinity. $\endgroup$ – John Doe Oct 7 '17 at 21:14
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    $\begingroup$ right, but I want to prove that f(x) converges to a finite or infinite value! $\endgroup$ – Tamir Oct 7 '17 at 21:19

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