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 - Page 102: 7


$\lim\limits_{x\to 8}(1+\sqrt[3] x)(2-6x^2+x^3)=390$

Work Step by Step

$\lim\limits_{x\to 8}(1+\sqrt[3] x)(2-6x^2+x^3)$ $=\lim\limits_{x\to 8}(1+\sqrt[3]x)\times\lim\limits_{x\to 8}(2-6x^2+x^3)$ (product law) $=[\lim\limits_{x\to 8}1+\lim\limits_{x\to 8}(\sqrt[3]x)]\times(\lim\limits_{x\to 8}2-\lim\limits_{x\to 8}6x^2+\lim\limits_{x\to 8}x^3)$ (sum and difference law) $=[\lim\limits_{x\to 8}1+\sqrt[3]{(\lim\limits_{x\to 8}x)}]\times(\lim\limits_{x\to 8}2-6\lim\limits_{x\to 8}x^2+\lim\limits_{x\to 8}x^3)$ (root law and constant multiple law) $=[1+\sqrt[3]8]\times(2-6\times8^2+8^3)$ $=390$
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