Finding the limit $f(2x)-f(x)$

Question

Let $f:(0,+\infty) \rightarrow \mathbb{R}$ differentiable such that $\left | f'(x) \right | \leq \frac{1}{x^2} ,$ for every $x>0$

Prove that $\lim_{x\rightarrow\infty} \left |f(2x)-f(x) \right|=0$

Here's my attempt:

$$\lim_{x\rightarrow\infty}\left |f(2x)-f(x) \right|= \lim_{x\rightarrow\infty}\frac{\left |f(2x)-f(x) \right|\cdot(2x-x)}{2x-x}=\left |f'(x) \right|\cdot x\leq \frac{1}{x^2}\cdot x=\frac{1}{x}$$

Now using the squeeze theorem we get that the limit is $0$

I know though that there's something wrong with my solution..

What do you think?

Answer 1

The key here is mean value theorem. We have $$|f(2x)-f(x)|=|xf'(\xi)|=x|f'(\xi)|\leq\frac{x}{\xi^2}$$ where $\xi$ is some number between $x$ and $2x$. Since $\xi>x$ it follows that $x/\xi^2<1/x$ and therefore $$|f(2x)-f(x)|<\frac{1}{x}$$ for all $x>0$. By squeeze theorem the desired conclusion follows immediately.

Answer 2

$$ \begin{split} | f(2x)-f(x)| &= \bigg| \int_x^{2x} f^\prime(u)du\bigg| \\ &\leq \int_x^{2x} |f^\prime(u)|du \\ &\leq \int_x^{2x} \frac {du}{u^2}\\ &\leq \frac 1 {2x} \end{split}$$ So the limit is 0.