Calculus: Early Transcendentals (2nd Edition)

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

Chapter 2 - Limits - Review Exercises: 28

Answer

$-\infty $

Work Step by Step

$\lim _{u\rightarrow 0^+}\dfrac {u-1}{\sin u}=\dfrac {0-1}{\sin 0^{+}}=\dfrac {-1}{0^{+}}=-\infty $
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