Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.7 L'Hopital's Rule - 4.7 Exercises: 88

Answer

The solution is $$\lim_{x\to\pi/2}(\pi-2x)\tan x=2.$$

Work Step by Step

To solve this limit follow the steps below. LR will stand for Apply L'Hopital's rule $$\lim_{x\to\pi/2}(\pi-2x)\tan x= \lim_{x\to\pi/2}\frac{\pi - 2x}{\cot x}= \left[\frac{\pi-2\frac{\pi}{2}}{\cot\frac{\pi}{2}}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to\pi/2}\frac{(\pi-2x)'}{(\cot x)'}=\lim_{x\to\pi/2}\frac{-2}{-\frac{1}{\sin^2 x}}=\lim_{x\to\pi/2}2\sin^2 x=2\sin^2\pi/2 = 2\cdot1^2=2.$$
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