## Calculus: Early Transcendentals 8th Edition

$2$
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 $2$.