Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.5 Derivatives of Trigonometric Functions - 3.5 Exercises: 16

Answer

\[ = \frac{1}{2}\]

Work Step by Step

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