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I'm trying to show that if $f:\mathbb{R}\to\mathbb{R}$ is continuously differentiable such that $\lim_{x\to\infty} \dfrac{f(x)}{x}=0$ and $\alpha=\lim_{x\to\infty} f'(x)$ exists and is finite, then $\alpha=0$. My idea is to implement the mean value theorem but I cannot do it.

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Hint

By MVT, there is $c_x\in (x,2x)$ s.t. $$\frac{f(2x)-f(x)}{x}=f'(c_x).$$

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  • $\begingroup$ Ohhh that's right! Thanks! $\endgroup$ – Habagat Maliksi May 4 '18 at 13:12
  • $\begingroup$ you are welcome :-) $\endgroup$ – Surb May 4 '18 at 13:13
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Suppose that $\alpha>0$. Then there's a $M>0$ such that $x\geqslant M\implies f'(x)>\frac\alpha2$. Therefore, if $x>M$,$$\frac{f(x)-f(M)}{x-M}>\frac\alpha2,$$by the mean value theorem. So$$\frac{f(x)-f(M)}{x-M}-\frac{f(x)}x>\frac\alpha4$$if $x$ is large enough. But this is impossible, since$$\frac{f(x)-f(M)}{x-M}-\frac{f(x)}x=\frac{Mf(x)-xf(M)}{x(x-M)}=\frac M{x-M}\cdot\frac{f(x)}x-\frac{f(M)}{x-M}\to0.$$

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