Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have the following function:

$$f(x,y):=\begin{cases}\frac{x^3y}{x^6+y^2}&,\;\;(x,y)\neq (0,0)\\{}\\0&,\;\;(x,y)=(0,0)\end{cases}$$

I'm asked about continuity at the origin and the limit of function there. Now, the limit doesn't exist since

$$\begin{align*}y=x^3&\implies f(x,x^3)=\frac{x^6}{x^6+x^6}=\frac12\xrightarrow [x\to 0]{}\frac12\;,\;\;\text{whereas}\\y=x&\implies f(x,x)=\frac{x^4}{x^6+x^2}=\frac{x^2}{x^4+1}\xrightarrow[x\to 0]{}0\end{align*}$$

My problem is: if I try to apply what's been shown in several questions in this site, namely polar coordinates, I get

$$\begin{cases}x=r\cos t\\y=r\sin t\end{cases}\implies f(r,t)=\frac{r^2\cos^3t\sin t}{r^4\cos^6t+\sin^2t}$$

and now I argue: if $\;\sin t=0\;$ then $\;x=0\;$ and clearly $\;f(0,y)=0\;$ , otherwise

$$\lim_{r\to 0}\frac{r^2\cos^3t\sin t}{r^4\cos^6t+\sin^2t}=\frac 0{0+\sin t}=0$$

and thus the limit is zero...where am I going wrong?! Thanks.

share|improve this question
You missed one $r$ in$f(r,t)$ –  Claude Leibovici Feb 1 at 10:04
Where, @ClaudeLeibovici ? It is $\;r^4\;$ in the numerator but it cancels out to $\;r^2\;$ with the denominator... –  Timbuc Feb 1 at 10:05
Sorry ! I feel just stupid. –  Claude Leibovici Feb 1 at 10:08

2 Answers 2

up vote 3 down vote accepted

Parametrizing the function in polar coordinates doesn't change the fact that in order for the limit to exist, for any $\epsilon > 0$, there must exist a $\delta > 0$ such that $|f(x,y) - L| < \epsilon$ whenever $|(x,y)| < \delta$. That is to say, $f$ can be made arbitrarily close to some fixed $L$ for any sufficiently small neighborhood of $(0,0)$. If you parametrize the function in polar coordinates, you can still have trajectories with $r \to 0^+$ for which the angle $t$ varies (and it may even vary as a function of $r$). If you fix $t$ to any particular value, then you are only looking at trajectories that proceed along a ray to the origin. As you observed with the trajectory $y = x^3$, the limit is not zero along this curved path.

enter image description here

Here is a plot of $f$. As you can see, curves of the form $y = cx^3$ for any $c \ne 0$ will give a nonzero limit. We can also show this with direct computation, with the parametrization $y = ct^3$, $x = t$. Then $$\lim_{(x,y) \to (0,0)} f(x,y) = \lim_{t \to 0^+} f(ct^3, t^3) = \frac{c}{1+c^2}.$$ The range of this function of $c$ is clearly $-1/2 \le c \le 1/2$.

share|improve this answer
Thank you. But how does the difference of paths in the above example gets expressed? As far as I can see the limit is always zero no matter what the path (i.e., the coordinate $\;t\;$) is, and that's my problem in understanding here. –  Timbuc Feb 1 at 10:18
As I pointed out, every trajectory toward the origin, even curved ones, must have the same limit in order for the limit to exist. If you fix $t$ as constant and just take $r \to 0^+$, you are looking only at straight line paths toward the origin, and that is not sufficient, as the example $y = x^3$ shows. If you use polar coordinates, $t$ is allowed to be a function of $r$ as $r$ approaches $0$. The particular form of that function that is required to get a nonzero limit is complicated, but it doesn't mean it doesn't exist. –  heropup Feb 1 at 10:21
Oh, I think I see what you do about $\;r=r(t)\;$ being a function of $\;t\;$ or the other way around, @heropup....thanks, I shall mull on this. +1 –  Timbuc Feb 1 at 10:30

If the limit exists all of the paths you use should give the same result but you cannot use this to prove the existence of the limit. We usually use it to show the non existence of the limit.

If you want to actually get the limit you have to consider the general case as you do using the above substitution.

One more thing is that when $x$ and $y$ goe to zero $r$ goes to zero for every value of $t$, which does not happen in your case.

So the limit does not exist.

share|improve this answer
Thanks for the response, but I can't understand:as far as I understood (or thought I understood), the polar coordiantes are as general a case as can be with rectangular coordinates, and I keep on getting zero no matter what the polar coordinates are. –  DonAntonio Feb 1 at 10:13
you get $\frac{o}{\sin t}$ which is undermined for $t=0$ –  Semsem Feb 1 at 10:15
you can not do this pointwise it is a limit –  Semsem Feb 1 at 10:16
@Semsem, can't I do particular cases, say $\;\sin t=0\;$ ? I think I can... –  Timbuc Feb 1 at 10:17
you can do it to disprove , not to prove –  Semsem Feb 1 at 10:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.