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