Answer
$$-1$$
Work Step by Step
Since $|x|=-x $ when $ x<0$, then we have
\begin{align*}
\lim _{x \rightarrow 0^-} \frac{\sin x}{|x|}&= \lim _{x \rightarrow 0} \frac{\sin x}{-x}\\
&= (-1) \lim _{x \rightarrow 0} \frac{\sin x}{x}\\
&=-1.
\end{align*}
Where we used Theorem 2 -- that is, $\lim _{x\rightarrow 0}\frac{ \sin x}{ x}=1$ .
