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: 18

Answer

$\lim\limits_{h\to 0}\frac{(2+h)^3-8}{h}=12$

Work Step by Step

$\lim\limits_{h\to 0}\frac{(2+h)^3-8}{h}$ $=\lim\limits_{h\to 0}\frac{(8+h^3+12h+6h^2)-8}{h}$ [use $(a+b)^3=a^3+3a^2b+3ab^2+b^3$] $=\lim\limits_{h\to 0}\frac{h^3+6h^2+12h}{h}$ $=\lim\limits_{h\to 0}\frac{h(h^2+6h+12)}{h}$ (factorize the numerator) $=\lim\limits_{h\to 0}(h^2+6h+12)$ (divide both the numerator and denominator by $h$) $=(0^2+6\times0+12)$ $=12$
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