I saw this problem on my problem book:
IF $\lim\limits_{x\to0 }f(x)=0$ and $\lim\limits_{x \to 0 }\frac{f(2x)- f(x)}{x} =0$ prove that $\lim_{x \to 0} \frac{f(x)}{x}=0 $.
I tried to solve it but failed the only thing that I was able to prove its that if $\lim_{x \to 0} \frac{f(x)}{x}$ exist in $\overline {\mathbb{R}}$it it either $0$ or $\pm \infty$ and that it east to show $$\lim_{x \to 0} \frac{f(x)}{x}= \lim_{x \to 0} \frac{f(x)-f(2x)}{x} + 2\lim_{x \to 0} \frac{f(2x)}{2x}$$
i.e $L = 2L$
But I was not able to prove that the limit exist .
Since $\lim_{x \rightarrow 0} f(x) = 0$ you can write $f(x)$ as a telescoping sum $$f(x) = \sum_{n=0}^{\infty} (f(2^{-n} x) - f(2^{-n-1}x))$$ $$ = \sum_{n=0}^{\infty} (2^{-n - 1}x) {f(2^{-n} x) - f(2^{-n-1}x) \over 2^{-n - 1} x}$$ Thus one has $$|f(x)| \leq \sum_{n=0}^{\infty} 2^{-n - 1}|x| \bigg|{f(2^{-n} x) - f(2^{-n-1}x) \over 2^{-n - 1} x}\bigg|$$ $$\leq \bigg(\sum_{n=0}^{\infty} 2^{-n - 1}|x|\bigg)\sup_{n \geq 0} \bigg|{f(2^{-n} x) - f(2^{-n-1}x) \over 2^{-n - 1} x}\bigg|$$ $$= |x| \sup_{n \geq 0} \bigg|{f(2^{-n} x) - f(2^{-n-1}x) \over 2^{-n - 1} x}\bigg|$$ So we have $$\bigg|{f(x) \over x}\bigg| \leq \sup_{n \geq 0} \bigg|{f(2^{-n} x) - f(2^{-n-1}x) \over 2^{-n - 1} x}\bigg|$$ Now the fact that ${\displaystyle \lim\limits_{x \to 0 }\frac{f(2x)- f(x)}{x} =0}$ shows that the supremum here goes to zero as $x \to 0$ and we are done.