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If $f(x)$ is differentiable for $x>0$ and $$\lim_{x\to\infty}f'(x)=0,$$ then $$\lim_{x\to\infty}f(x)$$ exists finite.

Is that statement right or wrong?

egreg
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sam0101
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    There is a famous function whose derivative is $\frac 1 x$ on $\mathbb{R}^+$. – nicomezi Jun 29 '18 at 09:11
  • Yeah but I want to know if this statment always correct? – sam0101 Jun 29 '18 at 09:13
  • What is the anti-derivative of the example and its limit? – Lutz Lehmann Jun 29 '18 at 09:14
  • what you mean by anti derivative? – sam0101 Jun 29 '18 at 09:16
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    Dont you think you could edit the title of the question to make it more precise? If people are looking for an answer this is really generic. And there are a couple of typos also...And now that i see your profile you have several question like that. It is only a suggestion – user1868607 Jun 29 '18 at 09:16
  • It is a question from college to prove if this statement is right or wrong always,I posted the same question here – sam0101 Jun 29 '18 at 09:23
  • @sam0101 nicomezi is trying to show you a counter-example. The existence of a single counter-example should tell you enought to judge on the validity of your statement. – M. Winter Jun 29 '18 at 09:23
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    The converse i.e. if $\lim_{x \to \infty}f(x)$ is finite, then $\lim_{x \to \infty}f'(x) =0$ is correct, given that f(x) is everywhere differentiable on $\mathbb{R^+}$ – Shak Jun 29 '18 at 09:26
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    @LordKK No it is not. – nicomezi Jun 29 '18 at 09:27
  • @nicomezi I think it can be proved easily using mean value theorem. Can you give counter-example, if it not true. I may be wrong. – Shak Jun 29 '18 at 09:30
  • @nicomezi thanks, I get it now, which is lan(x) – sam0101 Jun 29 '18 at 09:31
  • @LordKK An everywhere differentiable counterexample is hard to write analitycally but it exists. Consider $f'$ made of smooth bumps functions centered at every integer being tighter as $x \to \infty$ with equal heights. If those bumps tighten fast enough, $f$ will have a finite limit while $f'$ does not have any limit. – nicomezi Jun 29 '18 at 09:33
  • @nicomezi I am assuming that limit of f(x) at $\infty$ does exist. Then using mean value theorem, one can show that limit of $f'(x)$ at $\infty$ is $0$. Only one thing could have devoided this possibility, when f'(x) itself would have been discontinuous, but here since limit of f'(x) exist, it is at least defined at $\infty$ – Shak Jun 29 '18 at 09:39
  • If you assume it exists, then yes it is correct. But your comment was not very clear on that point. @LordKK – nicomezi Jun 29 '18 at 09:42
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    @LordKK: A simple counterexample is $f(x)=\sin(x^2)/x$. – Hans Lundmark Jun 29 '18 at 13:56
  • @sam0101: It's not even true if you drop the word “finite” (i.e., allow $f'(x)$ to have the improper limit $+\infty$ or $-\infty$): https://math.stackexchange.com/questions/1131369/does-a-bounded-function-converge-if-its-derivative-tends-to-zero – Hans Lundmark Jun 29 '18 at 14:18
  • @HansLundmark Thanks. So, given $\lim_{x\to \infty} f(x) =L$ there can be two possibilities. Either $\lim_{x\to \infty} f'(x)$ doesn't exist, but if it exists, it will necessarily be equal to zero. I am confused whether this function in your example can be called differentiable on $\mathbb{R^+}$. Can we consider function differentiable, if its derivative exists everywhere, but not defined at $\infty$ – Shak Jun 29 '18 at 16:15
  • @LordKK: Yes, certainly. $\infty$ is not an element of the set $\mathbb{R}^+$. – Hans Lundmark Jun 29 '18 at 20:27

1 Answers1

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First of all, your limits are strange. You might like to write $$ \lim_{x\to\infty} f'(x)=0\text{ and }\lim_{x\to\infty} f(x)=L. $$

Next, we are not doing your homework. You should try it yourself.

The first try should be $f'(x)=x^{-n}$ for $n\in\mathbb N$. Obviously, $\lim_{x\to\infty} f'(x)=0$. How about $f$?

(Be carefull to check $n=1$ and $n>2$ as different cases)

Mundron Schmidt
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