limit involving trig functions I'm not sure how to solve this limit.
$$
\lim_{x\to 0} \frac{\tan 6x}{\sin 2x}
$$
After some rearranging I get this.
$$
\lim_{x\to 0} \frac{\sin 6x}{\cos 6x} \cdot \lim_{x\to0} \frac{1}{\sin 2x}
$$
I know the limit is 3, but how do you get there using algebra? What am I missing?

$$\frac{\tan{6x}}{\sin{2x}} = \frac{\tan{6x}}{6x}\cdot\frac{2x}{\sin{2x}}\cdot \frac{6}{2}$$
And 
$$\lim_{x\rightarrow 0} \frac{\tan{6x}}{6x} = 1$$
$$\lim_{x\rightarrow 0} \frac{2x}{\sin{2x}} = 1$$
I'll leave the rest to you.
