Answer
$$\frac{6}{5}$$
Work Step by Step
\begin{align*}
\lim _{x\rightarrow 0} \frac{ \sin 3x\sin 2x}{x\sin 5x}&=\lim _{x\rightarrow 0} \frac{6}{5} \frac{ \sin 3x }{3x}
\frac{ \sin 2x}{2x } \frac{ 5x}{\sin 5x}\\
&= \frac{6}{5} \lim _{3x\rightarrow 0} \frac{ \sin 3x }{3x}
\lim _{2x\rightarrow 0} \frac{ \sin 2x}{2x } \lim _{5x\rightarrow 0} \frac{ 5x}{\sin 5x} \\
&= \frac{6}{5}.
\end{align*}
Where we used Theorem 2 -- that is, $\lim _{x\rightarrow 0}\frac{ \sin x}{ x}=1. $
