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: 7

Answer

$ - 1$

Work Step by Step

$$\eqalign{ & \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {3, - 1} \right)} \frac{{{x^2}y - x{y^3}}}{{x - y + 2}} \cr & {\text{ }}f\left( {x,y} \right) = \frac{{{x^2}y - x{y^3}}}{{x - y + 2}}\cr &{\text{ is a rational function}}\cr &{\text{ and }}f\left( {3, - 1} \right) = 6 \cr & {\text{The denominator is not 0,}}\cr &{\text{therefore we can find the limit by direct }} \cr & {\text{substitution}}{\text{.}} \cr & {\text{Substitute 3 for }}x{\text{ and }} - 1{\text{ for }}y \cr & \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {3, - 1} \right)} \frac{{{x^2}y - x{y^3}}}{{x - y + 2}} = \frac{{{{\left( 3 \right)}^2}\left( { - 1} \right) - \left( 3 \right){{\left( { - 1} \right)}^3}}}{{\left( 3 \right) - \left( { - 1} \right) + 2}} \cr & \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {3, - 1} \right)} \frac{{{x^2}y - x{y^3}}}{{x - y + 2}} = \frac{{ - 6}}{6} \cr & \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {3, - 1} \right)} \frac{{{x^2}y - x{y^3}}}{{x - y + 2}} = - 1 \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.