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I encountered the following question, regarding the Mean value theorem: Let $f$ be some function such that it is differentiable the interval $(5, \infty)$, and satisfies $\lim_{x\rightarrow\infty}f'(x)=0$. Prove that $\lim_{x\rightarrow\infty}f(2x)-f(x)=0$.

I am trying to use the limit definition (using $\varepsilon, M$) and the Mean value theorem, but I got stuck. Here is my attempt:

Let $\varepsilon>0$. For any $x>5$, By the Mean value theorem on $[x, 2x]$, there exists some $x<c<2x$ such that $f'(c)=\frac{f(2x)-f(x)}{x}$. I know that $f'(c)$ (and $|f'(c)|$) can be as small as I want, but I am stuck due to the $x$ in the denominator, and I don't know how to get rid of it.

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    $\begingroup$ It’s false: consider $f(x) = \log x$ $\endgroup$
    – fwd
    Mar 9 at 22:04

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This is not true. Let $f(x)=\sqrt{x}$. Then $f$ is differentiable and $f'(x)=1/(2\sqrt{x})$. This is tends to zero as $x\to\infty$. But $\sqrt{2x}-\sqrt{x}=\sqrt{x}(\sqrt{2}-1)\to+\infty$.

Maybe you want to solve another problem, that is $f(x+2)-f(x)\to 0$. This is indeed true as $f(x+2)-f(x)=2f'(c)$. And $f'(c)\to 0$.

Edit.

So we need to show that $f(x)/x\to 0$. For any $\varepsilon>0$ there is $T$ such that for $x\geq T$ we have $|f(x)'|<\varepsilon/2$. We apply intermediate value theorem for interval $[T,x]$. We get $|f(x)-f(T)|=|f'(c)(x-t)|\leq \varepsilon/2 |x|$. So $|f(x)/x|\leq |(f(x)-f(T))/x|+|f(T)/x|\leq \varepsilon/2+|f(T)|/x$. So for $x>2|f(T)|/\varepsilon$, we get $|f(x)/x|\leq \varepsilon$. This means that $|f(x)/x|\to 0$.

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  • $\begingroup$ Yes, the interval-of-difference needs to be bounded... $\endgroup$ Mar 9 at 22:10
  • $\begingroup$ And what about $\lim_{x\rightarrow\infty}\frac{f(x)}{x}=0$? How can I prove it? $\endgroup$
    – Dani
    Mar 10 at 19:33
  • $\begingroup$ I understand that for $T>0$ $|x-T|<|x|$, but how to get rid of $f(T)$ on the left side? $\endgroup$
    – Dani
    Mar 23 at 16:36
  • $\begingroup$ I edited accordingly $\endgroup$ Mar 25 at 9:14

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