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: 8


$\ln 2$

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, $\ln (1+1)=\ln 2$
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