<s>[INST] Can someone explain this trigonometric limit? I have
$$\lim \limits_{x\to 0}  \frac {\tan(2x)}{\sin(x)}$$ and in my case the result is $\frac{2}{1}$ =2 not whether it is right.
This is my procedure.
$$\lim \limits_{x\to 0}  \frac{\frac {\sin(2x)}{\cos(2x)}}{\frac{\sin(x)}{1}}= \lim \limits_{x\to 0}  {\dfrac {\sin(2x)}{(\cos(2x))(\sin(x))}}=\dfrac{2x\frac {\sin(2x)}{2x}}{\cos(2x)\frac{x\sin(x)}{x}}$$
I separate the limit.
$$\frac{\left(\lim \limits_{x\to 0}2x\right) \cdot \left(\lim \limits_{x\to 0}\frac {\sin(2x)}{2x}\right)}{\lim \limits_{x\to 0}\left(\cos(2x)\right)\cdot\left(\lim \limits_{x\to 0}\frac{x\sin(x)}{x}\right)} = \lim \limits_{x\to 0} \dfrac{2x}{x}=\frac{2}{1} =2$$
 [/INST] That does give you the correct answer, but it takes a bit more work than necessary.
An alternative method is to use the double-angle formula for $\sin$.  That is:
$$\sin(2x) = 2\cos(x)\sin(x)$$
Thus:
$$\begin{align}\require{cancel}
\lim_{x\to0} \frac{\tan(2x)}{\sin(x)} &= \lim_{x\to0} \frac{\sin(2x)}{\cos(2x)\sin(x)} \\
&= \lim_{x\to0} \frac{2\cos(x)\cancel{\sin(x)}}{\cos(2x)\cancel{\sin(x)}} \\
&= \lim_{x\to0} \frac{2\color{blue}{\cos(x)}}{\color{red}{\cos(2x)}} \\
&= \frac{2\cdot \color{blue}{1}}{\color{red}{1}} \\
&= \boxed2
\end{align}$$
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