$\displaystyle \lim_{x \to 0} ~\frac{sin(6x)}{sin(2x)cos(3x)}$

I can see that the limit of each component is zero but can't seem to find the logic behind the entire function being zero. How can I get it into a form where the denominator is not 0 and the limit of the denominator is not zero? The only manipulation I could see was to change the numerator to 2sin(3x)cos(3x) and cancel the cos(3x), but I am left with the same set of problems.

I remember seeing rules for division and multiplication of trig functions with different angles like cos(2x)/cos(x) = cos(2x-x). Did I just imagine this?

I can't seem to find anything on it; I can only find the standard trig identities (sum difference formulas etc.).

Thanks.

edit: cos(3x)*