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: 40

Answer

$\frac{4}{5}$

Work Step by Step

Factor each polynomial: $$\lim_{x\to 1}\frac{x^3-x^2+3x-3}{x^2+3x-4}\\=\lim_{x\to 1}\frac{(x-1)(x^2+3)}{(x-1)(x+4)}.$$ Cancel the common factors: $$\lim_{x\to 1}\frac{(x^2+3)}{(x+4)}\\=\frac{(1^2+3)}{(1+4)}\\=\frac{(1+4)}{5}\\=\frac{4}{5}$$
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