University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 13 - Practice Exercises - Page 749: 11

Answer

$\dfrac{1}{2}$

Work Step by Step

We need to find the limit for $\lim\limits_{(x,y) \to (1,1)} \dfrac{x-y}{x^2-y^2}$ or, $\lim\limits_{(x,y) \to (1,1)} \dfrac{x-y}{x^2-y^2}=\lim\limits_{(x,y) \to (1,1)} \dfrac{x-y}{(x-y)(x+y)}=\lim\limits_{(x,y) \to (1,1)} \dfrac{1}{x+y}$ Thus, $\dfrac{1}{1+1}=\dfrac{1}{2}$
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