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 - Page 169: 12

Answer

=0

Work Step by Step

\[\begin{gathered} \mathop {\lim }\limits_{\theta \, \to \,0} \,\,\frac{{{{\cos }^2}\theta - 1}}{\theta } \hfill \\ \hfill \\ factoring\,\,the\,\,numerator\, \hfill \\ \hfill \\ = \mathop {\lim }\limits_{\theta \, \to \,0} \,\,\,\,\,\frac{{\,\left( {\cos \,\,\theta + 1} \right)\,\left( {\cos \theta - 1} \right)}}{\theta } \hfill \\ \hfill \\ use\,\,the\,\,product\,\,of\,\,limits\,law \hfill \\ \hfill \\ = \,\left( {\mathop {\lim }\limits_{\theta \, \to \,0} \,\,\,\,\,\left( {\cos \,\,\theta + 1} \right)} \right)\,\left( {\mathop {\lim }\limits_{\theta \, \to \,0} \,\,\frac{{cos\theta - 1}}{\theta }} \right) \hfill \\ \hfill \\ from\,\,the\,\,theorem\,\,7.11 \hfill \\ \hfill \\ \mathop {\lim }\limits_{\theta \, \to \,0} \,\,\,\,\frac{{\cos \,\,\theta - 1}}{\theta } = 0\,\, \hfill \\ \hfill \\ then \hfill \\ \hfill \\ = \,\left( {1 + 1} \right)\,\left( 0 \right) \hfill \\ \hfill \\ = 0 \hfill \\ \hfill \\ \end{gathered} \]
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.