Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 2 - Limits - 2.3 Techniques for Computing Limits - 2.3 Exercises: 29

Answer

$3$

Work Step by Step

$\lim _{b\rightarrow 2}\dfrac {3b}{\sqrt {4b+1}-1}=\dfrac {3\times 2}{\sqrt {4\times 2+1}-1}=\dfrac {6}{\sqrt {9}-1}=\dfrac {6}{3-1}=3$
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