Answer
$ \frac{9}{7} $
Work Step by Step
\begin{align*}
\lim _{h\rightarrow 0} \frac{ \sin 9h}{\sin 7h}&=\lim _{h\rightarrow 0} \frac{9}{7} \frac{ \sin 9h}{ 9h} \frac{ 7h}{\sin 7h}\\
&= \frac{9}{7} \lim _{9h\rightarrow 0}\frac{ \sin 9h}{ 9h} \lim _{7h\rightarrow 0}\frac{ 7h}{\sin 7h}\\
&= \frac{9}{7} .
\end{align*}
Where we used Theorem 2 -- that is, $\lim _{x\rightarrow 0}\frac{ \sin x}{ x}=1. $
