Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 2 - Limits - Review Exercises - Page 124: 11


$\lim\limits_{x\to 1}{\frac{1-x^2}{x^2-8x+7}} = 3$

Work Step by Step

$\lim\limits_{x\to 1}{\frac{1-x^2}{x^2-8x+7}} = \lim\limits_{x\to 1}{\frac{(1-x)(1+x)}{(x-7)(x-1)}} =\lim\limits_{x\to 1}{\frac{-(x-1)(1+x)}{(x-7)(x-1)}} = \lim\limits_{x\to 1}{\frac{-(1+x)}{(x-7)}} =\frac{-(1+\lim\limits_{x\to 1}x)}{(\lim\limits_{x\to 1}x-7)} = \frac{-2}{-6}=3$
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