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: 6



Work Step by Step

We do what we have to according to the exercise: $$\lim_{x\to -1}(\frac{x^2+x+2}{x^2-9})=\frac{(-1)^2+(-1)+2}{(-1)^2-9}=\frac{-1}{4}.$$
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