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