Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Review - Exercises - Page 359: 14

Answer

$\lim\limits_{x \to (\pi/2)^-}~(tan~x)^{cos~x} = 1$

Work Step by Step

Let $~~~y = \lim\limits_{x \to (\pi/2)^-}~(tan~x)^{cos~x}$ $ln~y = \lim\limits_{x \to (\pi/2)^-}~ln~(tan~x)^{cos~x}$ $ln~y = \lim\limits_{x \to (\pi/2)^-}~cos~x~ln~(tan~x)$ $ln~y = \lim\limits_{x \to (\pi/2)^-}~\frac{ln~(tan~x)}{sec~x} = \frac{\infty}{\infty}$ We can use L'Hospital's Rule: $ln~y = \lim\limits_{x \to (\pi/2)^-}~\frac{ln~(tan~x)}{sec~x}$ $ln~y = \lim\limits_{x \to (\pi/2)^-}~\frac{\frac{cos~x}{sin~x}~\cdot~\frac{1}{cos^2~x}}{\frac{sin~x}{cos^2~x}}$ $ln~y = \lim\limits_{x \to (\pi/2)^-}~\frac{cos~x}{sin^2~x}$ $ln~y=\frac{0}{1}$ $ln~y= 0$ $y = e^0$ $y = 1$
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