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: 17

Answer

$\lim\limits_{h\to 0}\frac{(-5+h)^2-25}{h}=-10$

Work Step by Step

$\lim\limits_{h\to 0}\frac{(-5+h)^2-25}{h}$ $=\lim\limits_{h\to 0}\frac{(25+h^2-10h)-25}{h}$ (we use $(a-b)^2=a^2-2ab+b^2$) $=\lim\limits_{h\to 0}\frac{h^2-10h}{h}$ $=\lim\limits_{h\to 0}\frac{h(h-10)}{h}$ (factorize the numerator) $=\lim\limits_{h\to 0}(h-10)$ (divide both the numerator and denominator by $h$) $=0-10$ $=-10$
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