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: 72

Answer

$-2$

Work Step by Step

$$\lim _{x\rightarrow \frac {3\pi }{2}}\dfrac {\sin ^{2}x+6\sin x+5}{\sin ^{2}x-1}=\lim _{x\rightarrow \frac {3\pi }{2}}\dfrac {\left( \sin x+1\right) \left( \sin x+5\right) }{\left( \sin x+1\right) \left( \sin x-1\right) }=\lim _{x\rightarrow \frac {3\pi }{2}}\dfrac {\sin x+5}{\sin x-1}=\dfrac {\sin \dfrac {3\pi }{2}+5}{\sin \dfrac {3\pi }{2}-1}=\dfrac {-1+5}{-1-1}=\dfrac {4}{-2}=-2 $$
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