Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 14 - Section 14.2 - Limits and Continuity - 14.2 Exercise - Page 960: 19

Answer

$125$

Work Step by Step

$$\eqalign{ & \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( { - 1, - 2} \right)} {\left( {{x^2}y - x{y^2} + 3} \right)^3} \cr & {\text{Evaluating the limit by direct substitution}}{\text{.}} \cr & {\text{Substitute }} - {\text{1 for }}x{\text{ and }} - 2{\text{ for }}y \cr & \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( { - 1, - 2} \right)} {\left( {{x^2}y - x{y^2} + 3} \right)^3} = {\left( {{{\left( { - 1} \right)}^2}\left( { - 2} \right) - \left( { - 1} \right){{\left( { - 2} \right)}^2} + 3} \right)^3} \cr & {\text{Simplifying}} \cr & \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( { - 1, - 2} \right)} {\left( {{x^2}y - x{y^2} + 3} \right)^3} = {\left( { - 2 + 4 + 3} \right)^3} \cr & \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( { - 1, - 2} \right)} {\left( {{x^2}y - x{y^2} + 3} \right)^3} = 125 \cr} $$
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