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: 32

Answer

$=\tan 4 \approx 1.158$

Work Step by Step

$\lim _{t\rightarrow 4} \tan\dfrac {t-4}{\sqrt {t}-2}=\tan\dfrac {\left( \sqrt {t}\right) ^{2}-22}{\sqrt {t}-2}=\tan\dfrac {\left( \sqrt {t}-2\right) \left( \sqrt {t}+2\right) }{\left( \sqrt {t}-2\right) }=\tan(\sqrt {t}+2)=\tan\sqrt ( {4}+2)=\tan 4 \approx 1.158$
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