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: 44

Answer

The solution is: $$\lim_{x\to\pi/2}\frac{2\tan x}{\sec^2 x}=0.$$

Work Step by Step

To solve this limit follow the steps below: $$\lim_{x\to\pi/2}\frac{2\tan x}{\sec^2 x}=\lim_{x\to\pi/2}\frac{2\frac{\sin x}{\cos x}}{\frac{1}{\cos^2 x}}=\lim_{x\to\pi/2}\frac{2\sin x\cos^2 x}{\cos x}=\lim_{x\to\pi/2}2\sin x\cos x=2\sin(\pi/2)\cos(\pi/2)=2\cdot1\cdot0=0.$$
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