Find the follwing limit if $f$ is differentiable If $f$ is differentiable at $x=0$ and $f'(0) =2$, then find $$\lim_{x\to 0} \frac{2f(x) - 2f(2x)}{x}.$$

I don't know how to start, a hint will be appreciated.
Hints:

$$\lim_{x\to 0} \frac{2f(x)-2f(2x)}{x}=\lim_{x\to 0} \frac{2f(x)-2f(0)+2f(0)-2f(2x)}{x}=\lim_{x\to 0} \frac{2f(x)-2f(0)}{x}+\lim_{x\to 0}\frac{2f(0)-2f(2x)}{x}=2f'(0)-4f'(0)$$
f'(0)=2 $$\\\lim_{x\to 0}\dfrac{2f(x)-2f(2x)}{x}=\frac{0}{0}$$so you can use l'hoptial rule 
$$\lim_{x\to 0}\dfrac{2f(x)-2f(2x)}{x}=\\
2\lim_{x\to 0}\dfrac{f'(x)\times1-f'(2x)\times 2}{1}=\\
2\dfrac{f'(0)\times1-f'(0)\times 2}{1}=\\2(2-4)=-4$$