# Calculate the limit $\lim_{x\rightarrow 0} \frac{\sin(\sin(2x))}{x}$

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!

• Not able or not allowed ?
– user65203
Commented Mar 21, 2020 at 16:05
• allowed. Sorry for my english Commented Mar 21, 2020 at 16:14
• Don't worry, @Dunck. A lot of native English users confuse the two, as well! Commented Mar 21, 2020 at 16:15
• No problem, but I needed to know. That makes different answers.
– user65203
Commented Mar 21, 2020 at 16:15

We know that $$\lim_{x \to 0} \frac{x}{\sin{x}} = 1$$, so
$$\lim_{x\rightarrow 0} \frac{\sin(\sin(2x))}{x}= 2\lim_{x\rightarrow 0} \frac{\sin(\sin(2x))}{\sin(2x)}\frac{\sin(2x)}{2x}=2$$
• Please learn to format trig function: \tan x, \sin x, \cos x, \cot x,\sec x, \csc x` ... Commented Mar 21, 2020 at 16:12