Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - 1.6 Calculating Limits Using the Limit Laws - 1.6 Exercises: 22

Answer

$\frac{2}{3}$

Work Step by Step

$\lim\limits_{u \to 2}\frac{\sqrt {4u+1}-3}{u-2}=\lim\limits_{u \to 2}\frac{\sqrt {4u+1}-3}{u-2}*\frac{\sqrt {4u+1}+3}{\sqrt {4u+1}+3}=\lim\limits_{u \to 2}\frac{4u+1-9}{(u-2)(\sqrt {4u+1}+3)}=\lim\limits_{u \to 2}\frac{4u-8}{(u-2)(\sqrt {4u+1}+3)}=\lim\limits_{u \to 2}\frac{4(u-2)}{(u-2)(\sqrt {4u+1}+3)}=\lim\limits_{u \to 2}\frac{4}{\sqrt {4u+1}+3}=\frac{4}{\sqrt {4(2)+1}+3}=\frac{4}{6}=\frac{2}{3}$
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