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

Question

Consider the following 2-variable function:

$$f(x,y) = \frac{x^2y}{x^4+y^2}$$

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:

$$ x = r \cos \theta, \qquad y = r\sin\theta $$

Hence,

$$\lim_{(x,y) \to (0,0)} \frac{x^2y}{x^4 + y^2} = \lim_{r \to 0}\frac{r^2\cos^2\theta(r\sin\theta)}{r^4\cos^4\theta + r^2\sin^2\theta}$$

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 1

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);

Answer 2

You haven't taken into account what happens if $\theta$ is variable as a function of $r$ when $r$ goes to $0$. Choose $\theta$ so that $\sin \theta = r$, i.e. $\theta$ is approximately $r$ and you will get $\cos \theta$ is about 1 for small $r$, and then the limit will not be zero, so the limit doesn't exist.

If you want to use polar coordinates to show that a limit exists, particularly in the case where you want to show the limit is $0$ as $r \to 0$, then if you factor out a positive power of $r$ then you need to bound the remaining factor by either a constant or a multiple of a negative power of $r$ that is lower than the positive power you factored out. In your case you can't do this because when $\sin \theta = r$ you can't produce such a bound for the expression after you factor out $r$. If you had something like $r/(\cos^4 \theta + \sin^4 \theta)$ then you could bound $1/(\cos^4 \theta + \sin^4 \theta)$ by a constant for all $\theta$ and so you would then get that the limit is indeed $0$ as $r \to 0$.

Answer 3

Let $\alpha >0$, and consider the path $\gamma_\alpha(t) = (t,\alpha t^2)$. Then we have $f \circ \gamma_\alpha (t) = {\alpha t^4 \over t^4+ \alpha^2 t^4 }$, and the limit as $t \to 0$ is ${\alpha \over 1+\alpha^2}$ (in fact, it is constant along this path).

The limit exists along all of these paths, but is different. If the limit exists, its value must be independent of how $(x,y) \to 0$.

Answer 4

The OP asks:

1. What am I doing wrong with this method?

2. When should I not use polar coordinates to find limits of multivariable functions?

The answer to the second question is somewhat unsatisfactory: If you find a limit, then you can. If you don't find a limit, then you can't.

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So now, let's just leave that behind us and focus on the first question: "What am I doing wrong with this method?"

For this I will just consider the case where we have cartesian coordinates. The analogy with polar coordinates should be evident.

The mistake you made actually has nothing to do with "polar coordinates" per se, but with "limits". To this end, I'll first repeat the definition of the limit of a two-variable function here:

Suppose we have a function \begin{align}f:\mathbb R\times \mathbb R\supset U&\to \mathbb R\\ (x,y)&\mapsto f(x,y)\end{align} For a point $(a,b)\in\mathbb R$ we say that $\lim\limits_{(x,y)\to (a,b)}f(x,y)=L$, if and only if, $$\forall \varepsilon>0\,\exists \delta >0: \big(\Vert (x,y)-(a,b)\Vert<\delta \implies \vert f(x,y)-L\vert<\epsilon\big).\tag{*}$$ In words $(*)$ says that $f(x,y)$ will be close to $L$, whenever the point $(x,y)$ is sufficiently close to the point $(a,b)$.

Now comes your mistake: We have not defined what $\lim\limits_{x\to a}f(x,y)$ should mean. To say something about the limt of $f(x,y)$ we need to manipulate points in $\mathbb R^2$. But $x\to a$ means we are considering points in $\mathbb R$ which lie close to $a$ (which is definitely not a point in $\mathbb R^2$).

So this is a problem. If we would want to evaluate $\lim\limits_{x\to a}f(x,y)$, we would first have to define what this means. So let's do that:

Define $\ell_a\subset \mathbb R^2$ as the line $x=a$, i.e. $\ell_a =\left\{(x,y)\in\mathbb R^2\mid x=a\right\}$. Also introduce the notation: $d\big(\ell_a,(x,y)\big)=\text{distance between $(x,y)$ and $\ell_a$}$.

Now we say that $\lim\limits_{x\to a}f(x,y)=L(y)$, if and only if, $$\forall\varepsilon>0\,\exists\delta>0: \Big(d\big(\ell_a,(x,y)\big)\tag{**}<\delta \implies\vert f(x,y)-L(y)\vert<\varepsilon\Big).$$ In words $(**)$ says that $f(x,y)$ will be close to $L$, whenever the point $(x,y)$ is sufficiently close to the line $x=a$.

Notice the difference of the definitions in $(*)$ and $(**)$. The first tells us what happens if we are close to some point, the second tells us what happens if we are close to some line. Also, $(**)$ only says that we can get close to $L(y)$, which is some funtion of $y$. In general, proving that $\lim\limits_{x\to a}f(x,y)=L(y)$ is not at all easy and quite often not usesful.

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To sum up: The problem is that $\lim\limits_{r\to 0}\frac{r\cos^2\theta\sin\theta}{r^2\cos^4\theta+\sin^2\theta}$ is actually a rather stange and unuseful thing. If ever, it needs to be used with caution. This in particular means that it cannot be evaluated by simply substituting $r$ by $0$.

As an extra I would like to leave you with a funtion $f(r,\theta)$ for which $\lim\limits_{r\to 0}f(r,\theta)$ is of more use: $$\lim_{r\to 0}\frac {r\cos^2\theta \sin\theta}{r^2\cos^4\theta +1}=0\text{, because}$$

$$0<r<\delta\implies \left\vert\frac {r\cos^2\theta \sin\theta}{r^2\cos^4\theta +1}\right\vert<\left\vert\frac {r\cos^2\theta \sin\theta}{1}\right\vert=r\left\vert \cos^2\theta\sin\theta\right\vert<\delta\underbrace{\left\vert \cos^2\theta\sin\theta\right\vert}_{\text{bounded}}.\\\text{The singular cases where $\cos^2\theta \sin\theta=0$ are easily seen to be compatible.}$$

This rimes with its graph: