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Any hints for this question , My attempt;
Say $f(x):0$$\rightarrow$$\mathbb{R}$
The by MVT, there exists a $c$$\in$$(0,\infty)$ , such that;
$f'(c)=$$\frac{f(x)-f(0)}{x-0}$
but im not sure about this step..
$\lim_{c \to +\infty}$$f'(c)$=$\lim_{x \to +\infty}$$\frac{f(x)-f(0)}{x-0}$
I am quite sure there is a mistake, any hints would be appreciated.

If $\lim_{x \to +\infty}$$f(x)$ exists and is finite and $\lim_{x \to +\infty}$$f'(x)=b$ then $b=0$.

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marked as duplicate by Git Gud, mookid, egreg, Sujaan Kunalan, Davide Giraudo Mar 29 '14 at 18:56

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.

  • $\begingroup$ See this answer. $\endgroup$ – Git Gud Mar 29 '14 at 17:23
  • $\begingroup$ ok I agree, this might be a duplicate, what are the rules? should I delete this question? $\endgroup$ – otupygak Mar 29 '14 at 17:27
  • $\begingroup$ No, keep it. People will eventually close it as a duplicate. $\endgroup$ – Git Gud Mar 29 '14 at 17:28
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There is this $e^x$ trick. Consider $$f(x)=\frac{e^x f(x)}{e^x}$$ as $x\to\infty$. Use L'Hopital.

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$$f(x)=f(a)+\int_a^xf'(t)\ dt$$ So $\displaystyle\int_a^\infty f'(x)\ dx$ converges, because $\lim\limits_{x\to\infty}f(x)$ exists. Buf if $\lim\limits_{x\to\infty}f'(x)$ exists, then it must be $0$, else the integral wouldn't converge.

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  • $\begingroup$ the first line is only true when the derivative is a continuous function. $\endgroup$ – mookid Mar 29 '14 at 17:33
  • $\begingroup$ @mookid That's not true. $f$ only has to be differentiable. (this is of course the Newton integral) $\endgroup$ – user2345215 Mar 29 '14 at 17:34
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    $\begingroup$ I let you check here: "The Second part is somewhat stronger than the Corollary because it does not assume that f is continuous." $\endgroup$ – mookid Mar 29 '14 at 17:36
  • $\begingroup$ Another interesting point is the absolute continuity definition. $\endgroup$ – mookid Mar 29 '14 at 17:38
  • $\begingroup$ One nice counterexample is the Cantor function, for which $f(1) - f(0) = 1$ but $f'(x) = 0 $ almost everywhere. $\endgroup$ – mookid Mar 29 '14 at 17:40
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$$\limsup_{h\to 0} \left| \frac{f(x+h) - f(x)} h - b \right| = \limsup_{h\to 0} |f'(x+\theta_{x,h} h) - b| \le \limsup_{h\to 0} \sup_{0<\theta<1}|f'(x+\theta h) - b| =0 $$

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