Find $\lim_{x\to 0} \frac{\tan16x}{\sin2x}$ Find $\lim_{x\to 0} \frac{\tan16x}{\sin2x}$ 
I'm a little confused on limit trig.  Am i suppose to simplify tan or do I use the derivative quotient rule?
Please Help!!!

Hint: $$\frac{\tan 16x}{\sin 2x} = \frac{\sin 16x}{\cos 16 x}\frac{1}{\sin 2x} = \frac{\sin 16x}{16 x}\frac{1}{\cos 16 x}\frac{2x}{\sin 2x}\frac{16 x}{2x}= \frac{\sin 16x}{16 x}\frac{2x}{\sin 2x}\frac{8}{\cos 16 x}.$$
