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: 74

Answer

$\dfrac {1}{4}=0.25$

Work Step by Step

$$\lim _{\theta \rightarrow 0}\dfrac {\dfrac {1}{2+\sin \theta }-\dfrac {1}{2}}{\sin \theta }=\dfrac {\dfrac {2+\sin \theta -2}{2\times \left( 2+\sin \theta \right) }}{\sin \theta }=\dfrac {\dfrac {\sin \theta }{2\times \left( 2+\sin \theta \right) }}{\sin \theta }=\dfrac {1}{2\times \left( 2+\sin \theta \right) }=\dfrac {1}{2\times \left( 2+0\right) }=\dfrac {1}{4}=0.25$$
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