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