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I'm working on the problem:

Suppose $f(x)$ is differentiable on $(0,+\infty)$. If $$\lim_{x\rightarrow +\infty} \frac{f(x)}{x}=0,$$ show there is $x_n\rightarrow\infty$ such that $$\lim_{n\rightarrow\infty}f'(x_n)=0.$$

Here are some of my thoughts:

Let $n\geq 2$, then $\frac{f(n)-f(1)}{n-1}=f'(x_n)$ for some $x_n\in (1,n)$, so $f'(x_n)\rightarrow 0$. But I can't show $x_n\rightarrow +\infty.$

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The Mean Value Theorem says that for any $x$, there is a $\xi\in(x,2x)$ so that $$ \frac{f(2x)-f(x)}{x}=f'(\xi)\tag1 $$ Choose any $\epsilon\gt0$. Find an $M$ so that $x\ge M\implies\left|\frac{f(x)}{x}\right|\lt\frac{\epsilon}3$. Then for $x\ge M$, $$ \begin{align} \left|f'(\xi)\right| &=\left|\,\frac{f(2x)-f(x)}{x}\,\right|\\ &=\left|\,2\frac{f(2x)}{2x}-\frac{f(x)}x\,\right|\\ &\le2\left|\,\frac{f(2x)}{2x}\,\right|+\left|\,\frac{f(x)}x\,\right|\\[6pt] &\le\epsilon\tag2 \end{align} $$ Since $\epsilon\gt0$ was arbitrary, we can find a sequence $M_n\ge n$ which corresponds to $\epsilon=2^{-n}$, which gives a sequence $\xi_n\in(M_n,2M_n)$ so that $|f'(\xi_n)|\le2^{-n}$.

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  • $\begingroup$ robjohn.Nice answer. $\endgroup$ – Peter Szilas Nov 26 '19 at 9:55
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Suppose it is not true, there exists $c>0$ and $M$ such that $x>M$ implies that $|f'(x)|>c$, if $x>M, f(x)-f(M)=f'(y)(x-M)$ implies that $|{{f(x)}\over x}|$ $\geq c-|{{f(M)}\over x}|$ contradiction with the fact that $lim_{x\rightarrow+\infty}|{{f(x)}\over x}=0$.

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If the conclusion is not true then there exists $\epsilon >0$ and $M$ such that $|f'(x)| >\epsilon$ for all $x \geq M$. Using the fact that derivatives have IVP we see that we can actually make $f'(x) >\epsilon$ for all $x \geq M$ or $f'(x)<-\epsilon$ for all $x \geq M$. Consider the former case. Note that $f(n+1)-f(n) \geq \epsilon $ for all $n >M$ by MVT. Now you can see easily that $\frac {f(n)} n $ does not tend to $0$.

[If $n_0 >M$ then $f(n) \geq (n-n_0)\epsilon+f(n_0)$ for all $n >n_0$ so $\frac {f(n)} n \geq (1-\frac {n_0} n)\epsilon+\frac 1 nf(n_0) \to \epsilon$ as $n \to \infty$].

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