Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 15 - Differentiation in Several Variables - Chapter Review Exercises - Page 835: 14



Work Step by Step

Given$$\lim _{(x, y) \rightarrow(0,2)} \frac{\left(e^{x}-1\right)\left(e^{y}-1\right)}{x}$$ Since \begin{align*} \lim _{(x, y) \rightarrow(0,2)} \frac{\left(e^{x}-1\right)\left(e^{y}-1\right)}{x}&=\lim _{x \rightarrow0} \frac{\left(e^{x}-1\right) }{x}\lim _{y\rightarrow2} \left(e^{y}-1\right) \\ &= (e^2-1) \lim _{x \rightarrow0} \frac{\left(e^{x}-1\right) }{x}\\ & \text{using L'Hopital's Rule } \\ &=( e^2-1) \lim _{x \rightarrow0} \frac{\left(e^{x}\right) }{1}\\ &=e^2-1 \end{align*}
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