Answer
Given: $\lim\limits_{(x, y) \to (0, 0)}\frac{x^{2} + y^{2}} {{\sqrt {x^{2} + y^{2} + 1}}-1}$
Rationalize by multiply both numerator and denominator by $\sqrt {x^{2}+y^{2}+1}+1$
$=\lim\limits_{(x,y) \to (0,0)}\frac{x^{2}+y^{2}}{\sqrt {x^{2}+y^{2}+1}-1}\times \frac{\sqrt {x^{2}+y^{2}+1}+1}{\sqrt {x^{2}+y^{2}+1}+1}$
$=\lim\limits_{(x,y) \to (0,0)}{\sqrt {x^{2}+y^{2}+1}+1}$
We get intermediate form. Therefore, we will calculate the limit of the function in the following way.
To evaluate the limit along the x-axis use $y=0$
$f(x,0)={\sqrt {x^{2}+0^{2}+1}+1}=\sqrt {x^{2}+1}+1$
To evaluate limit along the y-axis; put $x=0$
$f(0,y)={\sqrt {0^{2}+y^{2}+1}+1}=\sqrt {y^{2}+1}+1$
For a limit to exist, all paths must converge to the same point. This is not the case, so the limit does not exist.
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 $2$.
