## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 15 - Differentiation in Several Variables - 15.2 Limits and Continuity in Several Variables - Exercises - Page 771: 7

#### Answer

$$\frac{1}{2}\frac{e^2-1}{e}$$

#### Work Step by Step

Since the function $\frac{e^{x^2}-e^{-y^2}}{x+y}$ is continuous at $(1,1)$, then we have $$\lim\limits_{(x,y) \to (1,1)} \frac{e^{x^2}-e^{-y^2}}{x+y}= \frac{e^{1}-e^{-1}}{1+1}=\frac{1}{2}\frac{e^2-1}{e}.$$

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