Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 13 - A Preview of Calculus: The Limit, Derivative, and Integral of a Function - Section 13.2 Algebra Techniques for Finding Limits - 13.2 Assess Your Understanding - Page 903: 14

Answer

$-54$

Work Step by Step

Use the limit properties: $(1) \ \lim\limits_{x \to a} l(x)=l (a) \\ (2) \lim\limits_{x \to a} k(x) l(x)=\lim\limits_{x \to a} k(x) \cdot \lim\limits_{x \to a} l(x)=k(a) \ l(a) $ where $a$ is a constant. $\lim\limits_{x \to -3} (2x^3)= \lim\limits_{x \to -3} (2) \cdot \lim\limits_{x \to -3} (x^3) \\=(2)(-3)^3\\=(2)(-27)\\=-54$
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.