TITLE: Can someone explain this trigonometric limit?
QUESTION [7 upvotes]: 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.
$$\frac{\left(\lim \limits_{x\to 0}2x\right) \cdot \left(\lim \limits_{x\to 0}\frac {\sin(2x)}{2x}\right)}{\lim \limits_{x\to 0}\left(\cos(2x)\right)\cdot\left(\lim \limits_{x\to 0}\frac{x\sin(x)}{x}\right)} = \lim \limits_{x\to 0} \dfrac{2x}{x}=\frac{2}{1} =2$$

REPLY [2 votes]: What are you allowed to use? An easier way of solving it is by expanding both numerator and denominator in Maclaurin series for $ x \to 0$: 
$$
\lim_{x \to 0}\frac{2x+O(x^3)}{x+O(x^3)}=2
$$