Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - 2.6 Trigonometric Limits - Exercises - Page 77: 34



Work Step by Step

\begin{align*} \lim _{x\rightarrow 0} \frac{ \tan 4x}{9x}&=\lim _{x\rightarrow 0} \frac{ \sin 4x}{9x} \frac{1}{\cos 4x} \\ &= \lim _{x\rightarrow 0} \frac{4}{9}\frac{ \sin 4x}{4x} \frac{1}{\cos 4x} \\ &= \frac{4}{9}\lim _{4x\rightarrow 0} \frac{ \sin 4x}{4x} \lim _{x\rightarrow 0} \frac{1}{\cos 4x}\\ &= \frac{4}{9}\frac{1}{\cos 0}\\ &=\frac{4}{9}. \end{align*} Where we used Theorem 2 -- that is, $\lim _{x\rightarrow 0}\frac{ \sin x}{ x}=1. $
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