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: 17


$\lim\limits_{x\to3}{\frac{x^4-81}{x-3}} =108$

Work Step by Step

$\lim\limits_{x\to3}{\frac{x^4-81}{(x-3)}} = \lim\limits_{x\to 3}{\frac{(x^2-9)(x^2+9)}{(x-3)}} =\lim\limits_{x\to 3}{\frac{(x-3)(x+3)(x^2+9)}{(x-3)}} =\lim\limits_{x\to 3}{\frac{(x+3)(x^2+9)}{1}}=(\lim\limits_{x\to 3}x+3)(\lim\limits_{x\to 3}x^2 +9)=6\times18=108$
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