Find the value of limit as x approaches 0

y =(sin3x) (cos2x) (sin2x) / (sin4x) (cos5x) (sin6x)

I am not sure how to approach this question. I haven't learnt the l'hopital's rule, so it'd be great if you could give a brief description, if it's needed to solve the question.
Thanks a lot!

Use the fact that lim(t→0) sin t/t = 1.

lim(x→0) [sin(3x) cos(2x) sin(2x)] / [sin(4x) cos(5x) sin(6x)]
= lim(x→0) [cos(2x)/cos(5x)] * [sin(3x) sin(2x)] / [sin(4x) sin(6x)]
= lim(x→0) (1/1) * [sin(3x) sin(2x)/x^2] / [sin(4x) sin(6x)/x^2]
= lim(x→0) [3 sin(3x)/(3x)] * [2 sin(2x)/(2x)] / {[4 sin(4x)/(4x)] * [6 sin(6x)/(6x)]}
= (3 * 2)/(4 * 6), using the fact above with t = 2x, 3x, 4x, 6x
= 1/4.

I hope this helps!