Calculate the following limit:
$$\lim_{x\rightarrow 0} \frac{\sin(\sin(2x))}{x}$$
I'm not allowed to use L'Hôpital's rule. I tried to make the substitutions $2x=y$ and $\sin(2x)=y$, but I couldn't get something to use $\lim_{x\rightarrow 0} \frac{\sin(x)}{x}$ (this is the only one I can use this moment). I tried to use $\sin(2x)=2\sin(x)\cos(x)$ as well, but I didn't get anywhere too. I'm only asking a hint this moment! Thanks!
We know that $\lim_{x \to 0} \frac{x}{\sin{x}} = 1$, so
$$\lim_{x\rightarrow 0} \frac{\sin(\sin(2x))}{x}= 2\lim_{x\rightarrow 0} \frac{\sin(\sin(2x))}{\sin(2x)}\frac{\sin(2x)}{2x}=2$$