Answer
\[ = \frac{1}{2}\]
Work Step by Step
\[\begin{gathered}
\mathop {\lim }\limits_{x \to \, - 3} \,\,\frac{{\sin \,\left( {x + 3} \right)}}{{{x^2} + 8x + 15}}\, \hfill \\
\hfill \\
factoring\,\,the\,\,denominator \hfill \\
\hfill \\
= \mathop {\lim }\limits_{x \to \, - 3} \,\,\frac{{\sin \,\left( {x + 2} \right)}}{{\,\left( {x + 5} \right)\,\left( {x + 3} \right)}} \hfill \\
\hfill \\
use\,\,the\,\,product\,\,rule\,\,for\,\,\,limits\, \hfill \\
\hfill \\
= \,\left( {\mathop {\lim }\limits_{x \to \, - 3} \,\,\frac{1}{{x + 5}}} \right)\,\left( {\mathop {\lim }\limits_{x \to \, - 3} \,\,\,\frac{{\sin \,\left( {x + 3} \right)}}{{x + 3}}} \right) \hfill \\
\hfill \\
evaluate \hfill \\
\hfill \\
= \,\left( {\frac{1}{{ - 3 + 5}}} \right)\,\left( 1 \right) \hfill \\
\hfill \\
= \frac{1}{2} \hfill \\
\end{gathered} \]
