I have come to conclusion that the most efficient and thorough way to prove whether or not a limit exists in three dimensions is to use polar coordinates.

$lim_{x,y \to (0,0)} \frac{x^3+y^3}{x^2+y^2}$

Substituting for polar coordinates:

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

$lim_{r \to 0^+} r (cos^3 \Theta + sin^3 \theta)$

As we can see r is independent of Theta and therefore the limit exists and is in fact 0. However showing that it is independent using notation is a little bit of a grey area.

$\vert sin^3 \Theta + cos^3 \Theta \vert$ $\le$ $1$ (Since sin and cos are bound by 1). Then $\vert r sin^3 \Theta + cos^3 \Theta \vert$ $\le \vert r \vert$ $r \to 0$. Is this logical? To be honest I am not even sure what this is saying, if cos and sin are bound by one, how does this have anything to do with r? Also if I had $2x^3$ in the numerator how would that change this? I am not even sure how polar coordinates would be substituted with the 2 I would assume the constant would be pulled outside.

  • $\begingroup$ It is often efficiently wrong too: math.stackexchange.com/questions/753381/… $\endgroup$
    – Git Gud
    Commented Jun 18, 2016 at 20:30
  • $\begingroup$ The case @GitGud mentions is different because in that case if you varied $\theta$ in an appropriate way as $r\to0$ you got different results. Here your argument is fine. Now matter how $\theta$ varies, you have established that $|f(x,y)|<|r|$. $\endgroup$
    – almagest
    Commented Jun 18, 2016 at 20:36
  • $\begingroup$ @almagest The given example doesn't fail for that reason, the limit is $0$ for all values of $\theta$. $\endgroup$
    – Git Gud
    Commented Jun 18, 2016 at 20:46
  • $\begingroup$ @GitGud I am not sure I understand. Are you saying you agree with me? $\endgroup$
    – almagest
    Commented Jun 18, 2016 at 20:50
  • $\begingroup$ @almagest I disagree. You claim that in the exampled I linked to one gets different results for different values of $\theta$. I'm saying this isn't true. $\endgroup$
    – Git Gud
    Commented Jun 18, 2016 at 20:55

2 Answers 2


This works in the particular case where the numerator and denominator of your function are both homogenous polynomials in $x$ and $y$ (and the one in the denominator has no nontrivial roots) -- that is, in every term the sum of the exponents if $x$ and $y$ is the same. In that case you have $$ \frac{p(x,y)}{q(x,y)} = r^k h(\theta) $$ where $h$ is some continuous function with period $2\pi$.

Then $h$ is necessarily bounded, and therefore you can conclude that $\lim_{x,y\to 0} \frac{p(x,y)}{q(x,y)} = 0$ if $k>0$.

However, beware that when the function doesn't split as nicely as this, note that it is not enough that the limit under $r\to 0^+$ exists separately for each $\theta$ -- even if the limit is the same for all $\theta$. For example, if $$ f(x,y) = \begin{cases} \frac{x^2+y^2}{y} & \text{when }y> 0 \\ 0 & \text{when }y\le 0 \end{cases} $$ In this case $\lim_{r\to 0^+} f(r,\theta)=0$ for each $\theta$ but $\lim_{x,y\to 0} f(x,y)$ nevertheless doesn't exist. The above reasoning breaks down because $h$ is now not continuous when $\theta$ is a multiple of $\pi$.

For another example where the function isn't even defined by cases, see Limit is found using polar coordinates but it is not supposed to exist., which Git Gud pointed out. (In that example the reasoning above breaks down even earlier; the dependence on $r$ and $\theta$ don't decouple at all).

  • $\begingroup$ I am not really convinced that polar coordinates are any more treacherous than Cartesian ones. The fact is you have to be careful whatever the coordinate system. $\endgroup$
    – almagest
    Commented Jun 18, 2016 at 21:04

Well,first of all, you really mean in 2 dimensions since the limit is defined by the domain,which lies in $\mathbb R^2$. The surface of the graph of the function lies in $\mathbb R^3$,but that's not relevant for the limit

Second, while r in this case is independent of $\theta$, I'm really not sure how that matters in the limit.$\vert sin^3 \Theta + cos^3 \Theta \vert$ $\le$ $1$ is correct regardless of whether or not the equations are separable and that's what matters in this particular limit.

What makes polar coordinates so useful in this particular example is that the equation is separable in polar coordinates and there's no clear way to do this in Cartesian coordinates. (Interestingly, if you did have $2x^3$ in the numerator, the 2 would in fact come out once you converted the coordinates. That's not in general true.)

An interesting historical note: The physicist Edwin Shroedinger, when he was devising his famous partial differential equation in $\mathbb R^3$ to describe quantum mechanical systems, was struggling with rearranging the equation. The domain and experimental data of the equation seemed to strongly suggest the equation was separable in it's variables, but he couldn't get it to separate no matter what he tried. In desperation,he turned to his friend, mathematician Hermann Weyl,and asked him to take a look at it.

He soon got a letter back from Weyl telling him the equation was separable, but only in polar coordinates!


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