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: 4

Answer

$\lim\limits_{x \to -1}(x^4-3x)(x^2+5x+3)=-4$

Work Step by Step

$\lim\limits_{x \to -1}(x^4-3x)(x^2+5x+3)$ $=\lim\limits_{x \to -1}(x^4-3x)\times\lim\limits_{x \to -1}(x^2+5x+3)$ (product law) $=[\lim\limits_{x \to -1}x^4-\lim\limits_{x \to -1}3x]\times[\lim\limits_{x \to -1}x^2+\lim\limits_{x \to -1}5x+\lim\limits_{x \to -1}3]$ (difference and addition law) $=[\lim\limits_{x \to -1}x^4-3\lim\limits_{x \to -1}x]\times[\lim\limits_{x \to -1}x^2+5\lim\limits_{x \to -1}x+\lim\limits_{x \to -1}3]$ (constant multiple law) $=[(-1)^4-3\times(-1)]\times[(-1)^2+5\times(-1)+3]$ $=-4$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.