University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 13 - Section 13.2 - Limits and Continuity in Higher Dimensions - Exercises - Page 691: 49

Answer

The limit does not exist

Work Step by Step

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