Question

\[\lim_{n \rightarrow 0} \ [\sin3x \sin2x\ ]\]

Answer 1

Its [sin3x/ sin2x]

Answer 2

This one requires a little trickery. There is an identity that says sinx/x=1 (sinofantything/anything=1. So you have work some magic to make it (sin3x/3x)/(sin2x/2x)

Answer 3

The limit of sinx over x as x approaches to 0 equals 1 is not an identity but a theorem.
Looking back to the question, the answer is 3/2.