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: 39



Work Step by Step

In order to simplify the given expression, we will use the following rules. $(a) \lim\limits_{x \to a} \dfrac{p(x)}{q(x)}=\dfrac{\lim\limits_{x \to a} p(x)}{\lim\limits_{x \to a} q(x)} \\ (b) \lim\limits_{x \to a} k(x)=k(a)$ ; where $a$ is a constant. --- Thus, we have: $$ \lim\limits_{x \to 2} \dfrac{(x^3-2x^2+4x-8)}{(x^2+x-6)}=\lim\limits_{x \to 2} \dfrac{(x-2)(x^2+4)}{(x-2)(x+3)} \\=\dfrac{\lim\limits_{x \to 2} (x^2+4)}{\lim\limits_{x \to 2} (x +3)}\\=\dfrac{(2)^2+4}{2+3} \\=\dfrac{4+4}{5} \\=\dfrac{8}{5}$$
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