Answer
limit does not exist.
Work Step by Step
We notice that if we directly substitute limits in the given function $f(x,y)=\frac{x^{4}-4y^{2}}{x^{2}+2y^{2}}$
Then $f(0,0)=\frac{0}{0}$
We get intermediate form. Therefore, we will calculate limit of function in the following way.
To evaluate the limit along the x-axis use $y=0$
$f(x,0)=\frac{x^{4}-4y^{2}}{x^{2}+2y^{2}}=\frac{x^{4}-0}{x^{2}+0}=x^{2}$
To evaluate limit along y-axis; put $x=0$
$f(0,y)=\frac{x^{4}-4y^{2}}{x^{2}+2y^{2}}=-2$
For a limit to exist, all paths must converge to the same point. This is not the case, so the limit does not exist.
