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: 47



Work Step by Step

$\lim _{x\rightarrow 2}\sqrt {\dfrac {4x+10}{2x-2}}=\sqrt {\dfrac {4\times 2+10}{2\times 2-2}}=\sqrt {\dfrac {18}{2}}=\sqrt {9}=3$
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