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I don't know how to prove the following statement. any help please..

if $f$ is differentiable function on $\mathbb{R}$ and $\lim_{x\rightarrow \infty }\frac{f(x)}{x}=1$, then $\lim_{x\rightarrow \infty }f'(x)=1$.

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  • $\begingroup$ @cansomeonehelpmeout $\lim_{x\rightarrow \infty }\frac{\sin x}{x}=0$ $\endgroup$ – sera Apr 4 at 10:12
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You cannot prove it, since it is not true. Take, for instance, $f(x)=x+\sin(x^2)$. Then you do have $\lim_{x\to\infty}\frac{f(x)}x=1$. However, $f'(x)=1+2x\cos(x^2)$, and so it is not true that $\lim_{x\to\infty}f'(x)=1$.

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  • $\begingroup$ Thank you so much. Can I ask more question? if limx→∞f′(x) exists, is the result the same? $\endgroup$ – purecj Apr 4 at 10:19
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    $\begingroup$ No. It it exists, it can only be $1$. This follows from the mean value theorem. $\endgroup$ – José Carlos Santos Apr 4 at 10:21
  • $\begingroup$ Thank you. If it exists, can I apply the l'Hôpital's rule? $\endgroup$ – purecj Apr 4 at 10:48
  • $\begingroup$ Yes. That will also work here. $\endgroup$ – José Carlos Santos Apr 4 at 10:50

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