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: 27



Work Step by Step

In order to simplify the given expression, we will use the following rules. (a) $\lim\limits_{x \to a} [k(x)]^n =[\lim\limits_{x \to a} k(x)]^n$ (b) $\lim\limits_{x \to a} k(x)=k(a)$ where $a$ is a constant. --- Thus, we have: $\lim\limits_{x \to 2} (3x-2)^{5/2}=(\lim\limits_{x \to 2} [(3x-2)^{5/2}]=[(3)(2)-2]^{5/2} =4^{5/2} =(2)^{5}=32$
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.