# Prove that $\lim_{x\rightarrow\infty}f(2x)-f(x)=0$ using the Mean value theorem

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 $$x such that $$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.

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

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 $$|f(x)/x|\to 0$$.

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