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: 21

Answer

$\frac{1}{6}$

Work Step by Step

$\lim\limits_{h \to 0}\frac{\sqrt {9+h}-3}{h}=\frac{\sqrt {9+(0)}-3}{(0)}=\frac{0}{0}=$indet so we have to do algebra $\lim\limits_{h \to 0}\frac{\sqrt {9+h}-3}{h}=\lim\limits_{h \to 0}\frac{\sqrt {9+h}-3}{h}*\frac{\sqrt {9+h}+3}{\sqrt {9+h}+3}=\lim\limits_{h \to 0}\frac{9+h-9}{(h)(\sqrt {9+h}+3)}=\lim\limits_{h \to 0}\frac{1}{\sqrt {9+h}+3}=\frac{1}{\sqrt {9+(0)}+3}=\frac{1}{6}$
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