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: 11

Answer

=5

Work Step by Step

\[\begin{gathered} \mathop {\lim }\limits_{x \to 0} \,\,\frac{{\,\,\tan \,5x}}{x} \hfill \\ \hfill \\ use\,\,the\,\,identity\,\,\tan \theta = \frac{{\sin \theta }}{{\cos \theta }} \hfill \\ \hfill \\ \tan \,5x = \frac{{\sin 5x}}{{\cos 5x}}\,, \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \mathop {\lim }\limits_{x \to 0} \,\,\frac{{\,\,\tan \,5x}}{x} = \mathop {\lim }\limits_{x \to 0} \,\frac{{\sin 5x}}{{x\cos 5x}} \hfill \\ \hfill \\ Use\,\,product\,\,of\,\,\,limits\, \hfill \\ \hfill \\ = 5\,\left( {\mathop {\lim }\limits_{x \to 0} \frac{1}{{\cos x}}} \right)\,\left( {\mathop {\lim }\limits_{x \to 0} \frac{{\sin 5x}}{{5x}}} \right) \hfill \\ \hfill \\ where\,\,\mathop {\lim }\limits_{x \to 0} \frac{{\sin \,5x}}{{5x}} = 1\, \hfill \\ \hfill \\ = 5\,\left( 1 \right)\,\left( 1 \right) \hfill \\ \hfill \\ = 5 \hfill \\ \end{gathered} \]
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