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: 25

Answer

$2$

Work Step by Step

Given: $\lim\limits_{(x,y) \to (0,0)}f(x,y)=\frac{x^{2}+y^{2}}{\sqrt {x^{2}+y^{2}+1}-1}$ Multiply both the numerator and denominator by the conjugate of $\sqrt {x^{2}+y^{2}+1}-1$ i.e. $\sqrt {x^{2}+y^{2}+1}+1$. $=\lim\limits_{(x,y) \to (0,0)}\frac{x^{2}+y^{2}}{\sqrt {x^{2}+y^{2}+1}-1}\times \frac{\sqrt {x^{2}+y^{2}+1}+1}{\sqrt {x^{2}+y^{2}+1}+1}$ $=\lim\limits_{(x,y) \to (0,0)}{\sqrt {x^{2}+y^{2}+1}+1}$ $={\sqrt {0^{2}+0^{2}+1}+1}$ $=1+1$ $=2$ Hence, the limit converges to $2$.
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