Answer
$$\frac{-8}{7}.$$
Work Step by Step
Since the function $ \frac{2x^2}{4x+y}$ is continuous at $(-2,1)$, then we have
$$ \lim\limits_{(x,y) \to (-2,1)}\frac{2x^2}{4x+y}=\frac{2(-2)^2}{4(-2)+1}=-\frac{8}{7}.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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