Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.2 - Limits and Continuity - 14.2 Exercise: 6

Answer

The limit approaches $-\frac{2}{3}$

Work Step by Step

No issues arise if we evaluate the function at the given point, so the limit approaches $\frac{(2)^2(-1) + 2(-1)^2}{2^2 - (-1)^2} = -\frac{2}{3}$
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