## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 14: Partial Derivatives - Section 14.2 - Limits and Continuity in Higher Dimensions - Exercises 14.2 - Page 796: 43

#### Answer

Limit does not exist

#### Work Step by Step

Consider a approach : $(x,y) \to (0,0)$ along $y=kx^2$ Then, we get $\lim\limits_{x \to 0}\dfrac{x^4-(kx)^2}{x^4+(kx)^2}=\lim\limits_{x \to 0}\dfrac{1-k^2}{1+k^2}$ or, $\lim\limits_{x \to 0}\dfrac{1-k^2}{1+k^2}=\dfrac{1-k^2}{1+k^2}$ This shows that there are multiple limit values, thus, the limit does not exist at the point (0,0) for the function $f(x,y)=\dfrac{x^4-y^2}{x^4+y^2}$.

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