I need to find the limit as x approaches 0 for tan 6x / sin 2x

Try L'Hopital's Rule.

Here's a way without L'Hopital.

\lim_{x\to\0}\frac{tan(6x)}{sin(2x)}

\lim_{x\to\0}\frac{sin(6x)}{cos(6x)}\cdot\frac{1}{ sin(2x)}

\lim_{x\to\0}\frac{sin(6x)}{sin(2x)}\cdot\frac{1}{ cos(6x)}

\lim_{x\to\0}\frac{\frac{sin(6x)}{x}}{\frac{sin(2x )}{x}}\cdot\lim_{x\to\0}\frac{1}{cos(6x)}

Since \lim_{x\to\0}\frac{sin(x)}{x}=1:

\lim_{x\to\0}\frac{6\cdot\frac{sin(6x)}{6x}}{2\tim \frac{sin(2x)}{2x}}\cdot\lim_{x\to\0}\frac{1}{cos( 6x)}=\frac{6\cdot{1}}{5\cdot{1}}\cdot{1}=\fbox{3}