Answer
$-\infty $
Work Step by Step
$\lim _{u\rightarrow 0^+}\dfrac {u-1}{\sin u}=\dfrac {0-1}{\sin 0^{+}}=\dfrac {-1}{0^{+}}=-\infty $
Calculus: Early Transcendentals (2nd Edition)
by Briggs, Bill L.; Cochran, Lyle; Gillett, Bernard
Published by
Pearson
ISBN 10:
0321947347
ISBN 13:
978-0-32194-734-5
Chapter 2 - Limits - Review Exercises - Page 124: 28
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