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 - Page 70: 17


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

Work Step by Step

$\lim\limits_{h \to 0}\frac{(h-5)^2-25}{h}=\lim\limits_{h \to 0}\frac{h^2-10h+25-25}{h}=\lim\limits_{h \to 0}\frac{h^2-10h}{h}=\lim\limits_{h \to 0}\frac{h(h-10)}{h}=\lim\limits_{h \to 0}\frac{(h-10)}{1}=\lim\limits_{h \to 0}{(h-10)}=-10$
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