University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.2 - Limit of a Function and Limit Laws - Exercises: 14

Answer

$\lim\limits_{x \to -2}(x^{3} -2x^{2} +4x +8) = - 16$

Work Step by Step

$\lim\limits_{x \to -2}(x^{3} -2x^{2} +4x +8) = \lim\limits_{x \to -2}x^{3} - 2 \lim\limits_{x \to -2}x^{2} + 4\lim\limits_{x \to -2}x + \lim\limits_{x \to -2}8$ = $-2^{3} - 2(-2)^{2} + 4*(-2) + 8$ $= -8 - 8 -8 + 8 $ $ = -16$
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.