Proving that the function is differentiable by using limit. [duplicate]

$$\displaystyle \lim_{x\to 0}f(x)=0$$

$$\displaystyle \lim_{x\to 0}\frac{f(2x)-f(x)}{x}=0$$

Prove, that $$\lim_{x\to 0}\frac{f(x)}{x}=0$$

So, I have some ideas:

1. Do $\lim_{x\to 0}\frac{f(x+x)-f(x)}{x+x-x}=0$ and $\lim_{x\to 0}f(x)=0$ imply that $f'(0)$ exist and $f'(0)=0$? If yes, than $lim_{x\to 0}\frac{f(x)}{x}=f'(x)=0$. But I'm in doubts, because the definition of derivative include h and I want to get the difference between them.

2. Other way to look:

$$f(2x)-f(x)=\gamma(x)x$$ $$f(\frac{x}{2^{n-1}})-f(\frac{x}{2^{n}})=\gamma(\frac{x}{2^{n}})\frac{x}{2^{n}}$$ $$f(\frac{x}{2^{n-2}})-f(\frac{x}{2^{n-1}})=\gamma(\frac{x}{2^{n-1}})\frac{x}{2^{n-1}}$$ $$f(\frac{x}{2^{n-3}})-f(\frac{x}{2^{n-2}})=\gamma(\frac{x}{2^{n-2}})\frac{x}{2^{n-2}}$$ $$f(\frac{x}{2^{n-4}})-f(\frac{x}{2^{n-3}})=\gamma(\frac{x}{2^{n-3}})\frac{x}{2^{n-3}}$$ $$\dots$$ $$f(\frac{x}{2^{n-(n-1)-1}})-f(\frac{x}{2^{n-(n-1)}})=\gamma(\frac{x}{2^{n-(n-1)}})\frac{x}{2^{n-(n-1)}}=f(x)-f(\frac{x}{2})=\gamma(\frac{x}{2})\frac{x}{2}$$ Sum and get $f(x)-f(\frac{x}{2^n})=\sum_{k=0}^{n-1}\gamma(\frac{x}{2^{n-k}})\frac{x}{2^{n-k}}$ $f(x)=f(\frac{x}{2^n})+x\sum_{k=0}^{n-1}\gamma(\frac{x}{2^{n-k}})\frac{1}{2^{n-k}}$

Using OP's second technique, we have $$f(x) = f(x/2^{n}) + x\sum_{k = 0}^{n - 1}\frac{\gamma(x/2^{k + 1})}{2^{k + 1}}$$ where $\gamma(x) \to 0, f(x) \to 0$ as $x \to 0$. Letting $n \to \infty$ we get $$\frac{f(x)}{x} = \sum_{k = 0}^{\infty}\frac{\gamma(x/2^{k + 1})}{2^{k + 1}}$$ Also given any $\epsilon > 0$ we have a $\delta > 0$ such that $|\gamma(x)|< \epsilon$ whenever $0 < |x| < \delta$. Hence $|\gamma(x/2^{k + 1})| < \epsilon$ and then we can see that the infinite series on the right of above equation is absolutely convergent as absolute value of each term is bounded by $\epsilon/2^{k + 1}$ and $\sum_{k = 0}^{\infty}\epsilon/2^{k + 1} = \epsilon$. It follows that $$\left|\frac{f(x)}{x}\right| \leq \epsilon$$ whenever $0 < |x| < \delta$ and hence $\lim\limits_{x \to 0}\dfrac{f(x)}{x} = 0$ No extra conditions on $f$ are needed to establish the result.

I must say that OP had got over with the tricky part of the solution and only slight manipulations regarding infinite series and $\epsilon, \delta$ type arguments were needed.

For prime number $p$ and any nonzero integer $k$, define $f(\frac{k}{p})=\frac{1}{p}$. For irrational number $a$,$f(a)=0$. Then $f$ is well-defined on $\mathbb{R}$. You may check that it is a counterexample. What happens if we add the condition that $f$ is continuous? I will leave it as your exercise.