Question

Finding the limit with trig

Answer 1

\[\lim \frac{ \sin3xsin2x }{ xsin5x }\]

Answer 2

Would I use my trig identities?

Answer 3

x approaches what ?

Answer 4

I think I would break the limit of a product into the product of limits
and use the limit sin x / x = 1
\[ \lim \frac{ \sin3x \sin2x }{ x\sin5x } \\
=\lim 3\frac{\sin 3x}{3x} \cdot \lim 2\frac{\sin 2x}{2x} \cdot \lim \frac{1}{5} \frac{5x}{\sin 5x}
\]