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