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 - Page 103: 21


$\lim\limits_{h\to 0}\frac{\sqrt{9+h}-3}{h}=\frac{1}{6}$

Work Step by Step

$A=\lim\limits_{h\to 0}\frac{\sqrt{9+h}-3}{h}$ Multiply both numerator and denominator by $\sqrt{9+h}+3$, we have $A=\lim\limits_{h\to 0}\frac{(9+h)-9}{h(\sqrt{9+h}+3)}$ (We see that ($\sqrt{9+h}-3)(\sqrt{9+h}+3)=(9+h)-9$ since $(a-b)(a+b)=a^2-b^2$) $A=\lim\limits_{h\to 0}\frac{h}{h(\sqrt{9+h}+3)}$ $A=\lim\limits_{h\to 0}\frac{1}{\sqrt{9+h}+3}$ (divide numerator and denominator by $h$) $A=\frac{1}{\sqrt{9+0}+3}$ $A=\frac{1}{6}$
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