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: 15



Work Step by Step

Solve the limit $\lim\limits_{(x,y) \to (1,1)} \dfrac{(x-1)(y-2)}{x-1}$ $\lim\limits_{(x,y) \to (1,1)} \dfrac{(x-1)(y-2)}{(x-1)}=\lim\limits_{(x,y) \to (1,1)}(y-2)$ Thus, $1-2=-1$
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