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!!!

Recall the following limits:


*

*$\lim_{x \to 0}\dfrac{\sin(ax)}{x} = a$

*$\lim_{x \to 0}\dfrac{\tan(bx)}{x} = b$


Note that
$$\dfrac{\tan(16x)}{\sin(2x)} = \dfrac{\dfrac{\tan(16x)}{x}}{\dfrac{\sin(2x)}x}$$
Can you finish it off now?
