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? 

$\lim_{x\to 0}\frac{\tan3x}{\sin2x}=\lim_{x\to 0}\frac{\tan3x}{3x}\frac{2x}{\sin2x}\frac{3x}{2x}=\frac{3}{2}$
or by using L'Hôpital's rule
$\lim_{x\to 0}\frac{\tan3x}{\sin2x}=\lim_{x\to 0}\frac{3(1+\tan^2 3x)}{2\cos2x}=\frac{3(1+\tan^2 3(0))}{2\cos2(0)}=\frac{3}{2}$
