Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.2 - Limits and Continuity - 14.2 Exercise - Page 911: 29


Continuous at $R^{2}$

Work Step by Step

Given: $f(x,y)=\frac{xy}{1+e^{x-y}}$ The given function is defined for all values of x and y except at $1+e^{x-y}=0$ This implies $e^{x-y}=-1$ But the value of $e^{x-y}$ cannot be negative; it is always greater than zero. This implies that $e^{x-y}>0$. Hence, the function $f(x,y)=\frac{xy}{1+e^{x-y}}$ is continuous at $R^{2}$.
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