Answer
Given: $f(x,y)=\frac{xy}{\sqrt {x^{2}+y^{2}}}$
Then $f(0,0)=\frac{0}{0}$
To evaluate limit along the x-axis; put $y=0$
$f(x,0)=\frac{xy}{\sqrt {x^{2}+y^{2}}}=0$
To evaluate limit along y-axis; put $x=1$
$f(1,y)=\frac{xy}{\sqrt {x^{2}+y^{2}}}=0$
However, the obtained identical limits along the axes do not show that the given limit is 0.
Then, approach (0,0) along another line, $y=x$ and $x≠0$
$f(x,x)=\frac{xy}{\sqrt {x^{2}+y^{2}}}=\frac{x^{2}}{\sqrt {x^{2}+x^{2}}}=\frac{x^{2}}{\sqrt {2x^{2}}}=\frac{x}{\sqrt 2}$
For a limit to exist, all the paths must converge to the same point. Since f(x,y) has two different values along two different paths, it follows that the limit does not exist.
Work Step by Step
By direct substitution:
$f(0,0)=\frac{0}{sqrt(0)}$ is indeterminate
Along the x-axis; y=0
$f(x,0)=\frac{0}{sqrt(x^{2})}=0$
Along the y-axis;x=0
$f(0,y)=\frac{0}{sqrt(y^{2})}=0$
Therefore the limit exists