$$\lim\_{x\to 0}\frac{\sin{6x}}{\sin{2x}}$$

I have no idea at all on how to proceed. I am guessing there is some trig rule about manipulating these terms in some way but I can not find it in my notes.

I tried to make $\tan$ into $\dfrac\sin\cos$

$$\frac{\sin6x}{\cos6x} \times \frac{1}{\sin2x}$$

But this doesn't get me anywhere as far as I can tell.