Calculus: Early Transcendentals (2nd Edition)

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

Chapter 2 - Limits - Review Exercises - Page 124: 21



Work Step by Step

$\lim _{x\rightarrow \dfrac {\pi }{2}}\dfrac {\dfrac {1}{\sqrt {\sin x}}-1}{x+\dfrac {\pi }{2}}=\dfrac {\dfrac {1}{\sqrt {\sin \dfrac {\pi }{2}}}-1}{\dfrac {\pi }{2}+\dfrac {\pi }{2}}=\dfrac {\dfrac {1}{\sqrt {1}}-1}{\pi }=\dfrac {1-1}{\pi }=0$
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