Limit with definite integral 
Given
  $$f(x)=\int _0^x \dfrac{\sin t}{t} dt$$
  calculate
  $$\lim _{x \rightarrow 0} \dfrac{2f(x)-f(2x)}{x-f(x)}.$$

I applied L'Hospital's rule so now I have:
$$\lim _{x \rightarrow 0} 2\dfrac{\frac{\sin x}{x}-\frac{\sin 2x}{2x}}{1-\frac{\sin x}{x}}$$
Now I don't know how to proceed.

Use Taylor's expansion at order $3$ after L'Hospital:
$$ \frac{2\sin x-\sin 2x}{x-\sin x}=\frac{2x-\dfrac{x^3}3-2x+\cfrac{4x^3}3+o(x^3)}{x-x+\cfrac{x^3}6+o(x^3)}=\frac{x^3+o(x^3)}{\cfrac{x^3}6+o(x^3)}=\frac{1+o(1)}{\frac 16+o(1)}.$$
