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 $2$.
