calculuslimitsmultivariable-calculus
Find $\lim_{(x,y) \to 0} x \ln (xy) ?$
Problem for this problem is already solved here on MSE.
But can't we find the limit using polar coordinates.
As $\lim _{r \to 0} r \cos \theta \ln( r^{2} \cos \theta \sin \theta) = 0$ so limit should be zero.
But the limit of this function doesn't exist.
What is wrong with this method$?$
There is no limit. Choose $x>0, m>0$, and $y=\sqrt{mx}$, then ${2x \over x^2 +x + y^2} = {2x \over x^2+(1+m)x} = { 2 \over x+(1+m)}$, and $\lim_ {x \downarrow 0} { 2 \over x+(1+m)} = { 2 \over 1+m}$. Since $m>0$ was arbitrary, there is no limit of the original function.
Your friend's technique would need to allow $\theta$ to be arbitrary as well, otherwise the limit is only taken over radial lines.
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);
Best Answer
Let $t \in A:= \Bbb{R}\setminus\{2k\pi,2k\pi +\pi/2:k \in Z\}$
$x\log(xy)=x\log{x} +x\log{y}$ thus for $x=r\cos{t}$ and $y=r \sin{t}$
we have that $f(r,t) \to 0$ as $r \to 0$ using the fact that $r\log{r} \to 0$ as $r \to 0$
Since $t$ is arbitrary in $A$ we conclude that the limit exists.