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 - Chapter Review - Review Exercises - Page 906: 8

Answer

$\frac{6}{7}$

Work Step by Step

Factor each polynomial: $$\lim_{x\to -3}\frac{x^2-9}{x^2-x-12}\\=\lim_{x\to -3}\frac{(x+3)(x-3)}{(x-4)(x+3)}.$$ Cancel the common factors: $$\lim_{x\to -3}\frac{x-3}{x-4}\\=\frac{(-3)-3}{(-3)-4}\\=\frac{6}{7}$$
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