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Let $f$ be a function in $(0,\infty)$ such that $f'(x)$ exists. In addition, $\lim\limits_{x\to \infty} f'(x)=L$ (finite) and $f(n)=0$ for every $n \in \Bbb N$.

Prove that $\lim\limits_{x\to\infty} f'(x)=0$

I think Rolle's theorem might help here, but I don't know how to use it.

Any hints are most welcome.

Thanks!

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Hint: using Rolles can select a sequence $c_n$ where $n<c_n<n+1$ and $f'(c_n)=0.$ This follows since $f(n)=f(n+1)$ (both are zero by your assumption). So whatever the limit of $f'(x)$ is, there is a subseqence of that converging to $0.$

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    $\begingroup$ You mean $f'(c_n) = 0$. $\endgroup$ – Eduardo Longa Feb 7 '16 at 19:14
  • $\begingroup$ @EduardoLonga Yes, typed to fast. Thanks, will fix. $\endgroup$ – coffeemath Feb 7 '16 at 19:14
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Hint: Mean value theorem in $[n,n+1]$ says that there exists a $c_n\in [n,n+1]$ such that $$f'(c_n)=\frac{f(n+1)-f(n)}{n+1-n}=0$$ for every $n\in\mathbb N$.

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