Limit: $\lim_{x\to 0}\frac{\tan3x}{\sin2x}$ $\lim_{x\to 0}\frac{\tan3x}{\sin2x}$= $\lim_{x\to 0}\frac{\frac{\sin(3x)}{\cos(3x)}}{\sin2x}=\lim_{x\to 0}\frac{\sin3x}{1}\cdot\frac{1}{\cos(3x)}\cdot\frac{1}{\sin(2x)}$
From this point I am lost. I believe I can pull a 3 and 2 out but I am not sure how. Can someone give me detailed instructions for a person in Calculus 1? 

Hint: $\lim_{x\to 0} \frac{\sin x}{x} = \lim_{x\to 0} \frac{\tan x}{x} = 1$
Rewrite your expression as
$$\frac{\tan 3x}{x} \cdot \frac{x}{\sin 2x} = \frac 32 \cdot \frac{\tan 3x}{3x} \cdot \frac{2x}{\sin 2x}$$
See what to do now?
