Answer
$\ln 2$
Work Step by Step
Solve the limit $\lim\limits_{(x,y) \to (1,1)} \ln (1+x^2y^2)$
$\lim\limits_{(x,y) \to (1,1)} \ln (1+x^2y^2)=\ln (1+(1)^2(1)^2)$
Thus, $\ln (1+1)=\ln 2$
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: 8
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