Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 14: Partial Derivatives - Section 14.2 - Limits and Continuity in Higher Dimensions - Exercises 14.2 - Page 796: 23



Work Step by Step

Re-arrange the given equation as follows; $\lim\limits_{(x,y) \to (1,-1)}\dfrac{(x^3-y^3)}{(x+y)}=\lim\limits_{(x,y) \to (1,-1)}\dfrac{(x+y)(x^2-xy+y^2)}{(x+y)}$ $\lim\limits_{(x,y) \to (1,-1)}\dfrac{(x+y)(x^2-xy+y^2)}{(x+y)}=\lim\limits_{(x,y) \to (1,-1)}(x^2-xy+y^2)$ Thus, $\lim\limits_{(x,y) \to (1,-1)}(x^2-xy+y^2)=1-(1)(-1)+(-1)^2=3$
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