Limit x2yx4+y2\frac{x^2y}{x^4+y^2} is found using polar coordinates but it is not supposed to exist.

Consider the following 2-variable function:

I would like to find the limit of this function as $(x,y) \rightarrow (0,0)$.

I used polar coordinates instead of solving explicitly in $\mathbb R^2$, and it went as the following:

Hence,

This simplifies to,

Simplifying $r^3/r^2$, we finally get;

Now from the above, we find that as $r \to 0$ the limit is $0$.

I wanted to verify this answer so I checked on Wolfram Alpha. Explicitly without changing to polar coordinates, it said that the limit does not exist at $(0,0)$ and rightly so. Then how is it that with polar coordinates, the limit exists and is $0$? Am I doing something wrong in this method?

Also, what should I do in this situation, and when should I NOT use polar coordinates to find limits of multi-variable functions?

The limit is not defined because in order for the limit to exist, the value of the function for every possible path to $(0,0)$ must tend to the same finite value. When $y = x^2$, you have not necessarily shown that the limit is in fact $0$. When you transformed to polar coordinates and then took the limit as $r \to 0$, you are assuming that $\theta$ is a fixed constant. Therefore, you are looking only at paths that follow a straight line to the origin.

Mathematica code:

F[x_, y_] := x^2 y/(x^4 + y^2)
op = ParametricPlot3D[{r Cos[t], r Sin[t], F[r Cos[t], r Sin[t]]},
{r, 0, Sqrt[2.1]}, {t, -Pi, Pi}, PlotPoints -> 40, MaxRecursion -> 8,
Mesh -> {10, 48}, PlotRange -> {{-1, 1}, {-1, 1}, {-1/2, 1/2}},
SphericalRegion -> True, Axes -> False, Boxed -> False];
an = Show[op, ViewPoint -> {{Cos[2 Pi #], Sin[2 Pi #], 0}, {-Sin[2 Pi #],
Cos[2 Pi #], 0}, {0, 0, 1}}.{1.3, -2.4, 2}] & /@ (Range[40]/40);