Answer
$\dfrac{1}{2}$
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, $e^{0-\ln 2}=\dfrac{1}{2}$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
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
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