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: 46


Limit does not exist

Work Step by Step

Consider a approach : $(x,y) \to (0,0)$ along $y=mx; m\ne 1$ Then, we get $\lim\limits_{x \to 0}\dfrac{x^2-kx}{x-kx}=\lim\limits_{x \to 0}\dfrac{x(x-k)}{x(1-k)}=\dfrac{-k}{1-k}$ This shows that there are multiple limit values when the approach is different, so, the limit does not exist at the point (0,0) for the given function $f(x,y)=\dfrac{x^2-y}{x-y}$.
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