Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.3 - Calculating Limits Using the Limit Laws - 2.3 Exercises: 22

Answer

$\lim\limits_{u\to2}\frac{\sqrt{4u+1}-3}{u-2}=\frac{2}{3}$

Work Step by Step

$A=\lim\limits_{u\to2}\frac{\sqrt{4u+1}-3}{u-2}$ Multiply both numerator and denominator by $\sqrt{4u+1}+3$ We notice that $(\sqrt{4u+1}-3)(\sqrt{4u+1}+3)=(4u+1)-9=4u-8=4(u-2)$ because $(a-b)(a+b)=a^2-b^2$ Therefore, $A=\lim\limits_{u\to2}\frac{4(u-2)}{(u-2)(\sqrt{4u+1}+3)}$ (divide both numerator and denominator by $u-2$) $A=\lim\limits_{u\to2}\frac{4}{\sqrt{4u+1}+3}$ $A=\frac{4}{\sqrt{4\times2+1}+3}$ $A=\frac{2}{3}$
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