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: 75

Answer

$-\dfrac {1}{2}=-0.5$

Work Step by Step

$\lim _{x\rightarrow 0}\dfrac {\cos x-1}{\sin ^{2}x}=\dfrac {\cos x-1}{1-\cos ^{2}x}=\dfrac {-\left( 1-\cos x\right) }{\left( 1-\cos x\right) \left( 1+\cos x\right) }=-\dfrac {1}{1+\cos x}=-\dfrac {1}{1+\cos 0}=-\dfrac {1}{2}=-0.5$
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