Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 15 - Differentiation in Several Variables - 15.2 Limits and Continuity in Several Variables - Exercises - Page 771: 4



Work Step by Step

Since the function $ \frac{2x^2}{4x+y}$ is continuous at $(-2,1)$, then we have $$ \lim\limits_{(x,y) \to (-2,1)}\frac{2x^2}{4x+y}=\frac{2(-2)^2}{4(-2)+1}=-\frac{8}{7}.$$
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