Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 14 - A Preview of Calculus: The Limit, Derivative, and Integral of a Function - 14.2 Algebra Techniques for Finding Limits - 14.2 Assess Your Understanding - Page 884: 39



Work Step by Step

Factor each polynomial: $$\lim_{x\to 2}\frac{x^3-2x^2+4x-8}{x^2+x-6}\\=\lim_{x\to 2}\frac{(x-2)(x^2+4)}{(x-2)(x+3)}.$$ Cancel the common factors: $$\lim_{x\to 2}\frac{(x^2+4)}{(x+3)}\\=\frac{(2^2+4)}{(2+3)}\\=\frac{(4+4)}{5}\\=\frac{8}{5}$$
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