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:

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


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

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?

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    $\begingroup$ It's odd - this conflicts with what I was taught last semester, that converting to polar is a foolproof way of proving a limit exists. $\endgroup$
    – user66698
    Commented Apr 15, 2014 at 7:46
  • 1
    $\begingroup$ I think @heropup's answer is more than convincing :) $\endgroup$
    – mesllo
    Commented Apr 15, 2014 at 8:10
  • $\begingroup$ @MyGlasses Did you know that the SE search engine can't read TeX? Consequently TeXifying the title doesn't help that much. $\endgroup$ Commented Apr 14, 2017 at 10:14
  • $\begingroup$ @JyrkiLahtonen I think these edits I have done so far, help us to find related and also duplicate in right panel (related panel). But if my work about these edits is un-useful, I will stop it. tell me please. $\endgroup$
    – Nosrati
    Commented Apr 14, 2017 at 10:27
  • $\begingroup$ I think the most important point is that when $r$ goes to zero, it's not uniform about $\theta$. It's easy to prove that when $r$ goes to zero, the limit exists and is equal, Also, $\delta$ is only related to $\epsilon$, independent of $\theta$, then limit exists. $\endgroup$
    – DaNan
    Commented Jul 7 at 9:41

4 Answers 4


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.

enter image description here enter image description here

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);
  • 12
    $\begingroup$ $r$ might not depend on $\theta$, but $\theta$ is most certainly allowed to depend on $r$. For if not, then the only paths that can be considered as $r \to 0$ would be those paths that move in a straight line to $(0,0)$, and as the choice of $y = x^2$ shows, such straight-line paths do not suffice. If $\theta$ varies with $r$, the calculation of the limit is no longer necessarily $0$. $\endgroup$
    – heropup
    Commented Apr 14, 2014 at 14:46
  • 3
    $\begingroup$ My initial reasoning was that any value of theta will not matter since we have r multiplied with an expression containing all the theta values, and r is tending to zero. I thought that whatever there is in that expression will not matter since it will be zero regardless, because it is multiplied with r. $\endgroup$
    – mesllo
    Commented Apr 14, 2014 at 15:06
  • 3
    $\begingroup$ @James If you fix $\theta$, the path you're taking to the origin is a straight line. There are many more paths than straight lines! (Great pictures, by the way!) $\endgroup$
    – user98602
    Commented Apr 14, 2014 at 15:53
  • 2
    $\begingroup$ @James If you allow θ->0 as r->0, you see that the "expression containing all the theta values"->0/0. This presents a problem and you would need to resolve it to show that f->0. $\endgroup$ Commented Apr 14, 2014 at 17:56
  • 4
    $\begingroup$ The images were created in Mathematica 9. $\endgroup$
    – heropup
    Commented Apr 15, 2014 at 0:55

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 $r \to 0$.

  • 4
    $\begingroup$ can u pls give an example as to what u r trying to say $\endgroup$
    – godonichia
    Commented Nov 7, 2014 at 19:10

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 $(x,y) \to 0$.

  • $\begingroup$ @StubbornAtom: I made a mistake, thanks for catching it. $\endgroup$
    – copper.hat
    Commented Jan 24, 2019 at 21:57

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.

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.

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:

Graph of the aforementioned function

  • 1
    $\begingroup$ Great answer! Can you also help me this question? I still stuck in there. $\endgroup$
    – Eric
    Commented Dec 4, 2016 at 17:49

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