Evaluating $ \lim_{x\to0} \frac{\tan(2x)}{\sin x}$ $$\lim_{x\to0} \frac{\tan(2x)}{\sin x}$$
How would I evaluate that? I was thinking changing the tan to sin/cos, but when I tried that, it did not work.

Since $\displaystyle\tan(2x)=\frac{2\tan x}{1-\tan^2x}$
Set $t=\frac x2$
Using tangent half-angle formula we get;
$\tan x=\frac{2t}{1-t^2}$, $\quad$$\sin x=\frac{2t}{1+t^2}$
Hence $\displaystyle\lim\limits_{x\to 0}\frac{\tan(2x)}{\sin(x)}=\displaystyle\lim\limits_{x\to 0}\frac{2\tan x}{1-\tan^2x}\cdot\frac{1}{\sin x}$
$\displaystyle=\lim\limits_{t\to 0}\frac{2\big(\frac{2t}{1-t^2}\big)}{1-\big(\frac{2t}{1-t^2}\big)^2}\cdot\frac{1+t^2}{2t}=2\lim\limits_{t\to 0}\frac{1-t^4}{1+t^4-6t^2}=2\cdot 1=2$
