Calculus: Early Transcendentals (2nd Edition)

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

Chapter 2 - Limits - 2.6 Continuity - 2.6 Exercises: 76



Work Step by Step

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