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: 40

Answer

$$\frac{4}{9}.$$

Work Step by Step

\begin{align*} \lim _{x\rightarrow 0} \frac{ \tan 4x}{\tan 9x}&=\lim _{x\rightarrow 0} \frac{ \sin 4x}{\cos 9x} \frac{\cos 9x}{\sin 9x} \\ &= \lim _{x\rightarrow 0} \frac{4}{9}\frac{ \sin 4x}{4x} \frac{9x}{\sin 9x} \frac{\cos 9x}{\cos 4x} \\ &= \frac{4}{9}\lim _{4x\rightarrow 0} \frac{ \sin 4x}{4x} \lim _{9x\rightarrow 0}\frac{9x}{\sin 9x} \lim _{x\rightarrow 0}\frac{\cos 9x}{\cos 4x}\\ &= \frac{4}{9}\frac{\cos 0}{\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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