Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 14: Partial Derivatives - Section 14.2 - Limits and Continuity in Higher Dimensions - Exercises 14.2 - Page 796: 49


Limit does not exist

Work Step by Step

Consider our first approach : $(x,y) \to (1,1)$ along $x=1$ Then, we get $\lim\limits_{y \to 1}\dfrac{y^2-1}{y-1}=\lim\limits_{y \to 1}\dfrac{(y-1)(y+1)}{y-1}=2$ Next, consider our second approach : $(x,y) \to (1,1)$ along $x=y$ Then, we get $\lim\limits_{y \to 1}\dfrac{y^3-1}{y-1}=\lim\limits_{y \to 1}\dfrac{(y-1)(y^2+y+1)}{(y-1)}$ or, $\lim\limits_{y \to 1} (y^2+y+1)=(1^2+1+1)=3$ This shows that there are different limit values when the approach is different, so, the limit does not exist at the point (0,0) for the function $f(x,y)=\dfrac{xy^2-1}{y-1}$.
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.