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: 7



Work Step by Step

Solve the limit $\lim\limits_{(x,y) \to (0,\ln 2)} e^{x-y}$ $\lim\limits_{(x,y) \to (0,\ln 2)} e^{x-y}=e^{0-\ln 2}$ Thus, $e^{0-\ln 2}=\dfrac{1}{2}$
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