University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 13 - Section 13.2 - Limits and Continuity in Higher Dimensions - Exercises - Page 690: 1



Work Step by Step

Solve the limit $\lim\limits_{(x,y) \to (0,0)}\dfrac{3x^2-y^2+5}{x^2+y^2+2}$. Thus, we have $\lim\limits_{(x,y) \to (0,0)}\dfrac{3x^2-y^2+5}{x^2+y^2+2}=\dfrac{3(0)^2-(0)^2+5}{(0)^2+(0)^2+2}=\dfrac{5}{2}$
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