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: 18

Answer

12

Work Step by Step

$\lim\limits_{h \to 0}\frac{(2+h)^3-8}{h}=\frac{(2+(0))^3-8}{(0)}=\frac{0}{0}=$indet so we have to do algebra $\lim\limits_{h \to 0}\frac{(2+h)^3-8}{h}=\lim\limits_{h \to 0}\frac{h^3+6h^2+12h+8-8}{h}=\lim\limits_{h \to 0}h^2+6h+12=(0)^2+6(0)+12=12$
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