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: 19


$\lim\limits_{x\to81}{\frac{\sqrt[4] x - 3}{x-81}} =\frac{1}{108}$

Work Step by Step

$\lim\limits_{x\to81}{\frac{\sqrt[4] x - 3}{x-81}} =\lim\limits_{x\to81}{\frac{\sqrt[4] x -3}{(\sqrt x-9)(\sqrt x +9)}} = \lim\limits_{x\to81}{\frac{\sqrt[4] x -3}{(\sqrt[4] x-3)(\sqrt[4] x+3)(\sqrt x +9)}} =\lim\limits_{x\to81}{\frac{1}{(\sqrt[4]x +3)(\sqrt x +9)}} ={\frac{1}{(\lim\limits_{x\to81}\sqrt[4]x +3)(\lim\limits_{x\to81}\sqrt x +9)}} =\frac{1}{(\sqrt[4]{\lim\limits_{x\to81}x} +3)(\sqrt {\lim\limits_{x\to81}x} +9)} = \frac{1}{(\sqrt[4] {81} +3) (\sqrt {81}+9)}=\frac{1}{108}$
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