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: 42

Answer

$\frac{13}{28}$

Work Step by Step

Factor each polynomial: $$\lim_{x\to 3}\frac{x^3-3x^2+4x-12}{x^4-3x^3+x-3}\\=\lim_{x\to 3}\frac{(x-3)(x^2+4)}{(x-3)(x^3+1)}.$$ Cancel the common factors: $$\lim_{x\to 3}\frac{(x^2+4)}{(x^3+1)}\\=\frac{(3^2+4)}{(3^3+1)}\\=\frac{(9+4)}{27+1}\\=\frac{13}{28}$$
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