Consider the following 2-variable function:

f(x,y)=x2yx4+y2

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

I used polar coordinates instead of solving explicitly in R2, and it went as the following:

x=rcosθ,y=rsinθ

Hence,

lim

This simplifies to,

\lim_{r \to 0} \frac{r^3 \cos^2\theta\sin\theta}{r^2(r^2\cos^4\theta + \sin^2\theta)}

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

\lim_{r \to 0} \frac{r (\cos^2\theta\sin\theta)}{r^2\cos^4\theta + \sin^2\theta}

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?

**Answer**

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);
```

**Attribution***Source : Link , Question Author : mesllo , Answer Author : heropup*