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 691: 23

Answer

$3$

Work Step by Step

$\lim\limits_{(x,y) \to (1,-1)}\dfrac{(x^3-y^3)}{(x+y)}=\lim\limits_{(x,y) \to (1,-1)}\dfrac{(x+y)(x^2-xy+y^2)}{(x+y)}$ Thus, we get $\lim\limits_{(x,y) \to (1,-1)}\dfrac{(x+y)(x^2-xy+y^2)}{(x+y)}=\lim\limits_{(x,y) \to (1,-1)}(x^2-xy+y^2)$ Hence, $1-(1)(-1)+(-1)^2=3$
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