## Calculus 8th Edition

Published by Cengage

# Chapter 14 - Partial Derivatives - 14.2 Limits and Continuity - 14.2 Exercises - Page 950: 17

#### 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 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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