1
$\begingroup$

Study the continuity of $$f(x,y)=\begin{cases}\dfrac{xy+y^2}{x^4+y^2}&\text{if }(x,y)\neq(0,0),\\0&\text{if }(x,y)=(0,0),\end{cases}$$ at $(x,y)=(0,0)$ using polar coordinates.


I know that $f(0,0)=0$ so, if $\lim_{(x,y)\to(0,0)}{f(x,y)}$ exists then it must be equal to $0$.

I want to prove that is not continuous at origin using polar coordinates. Let $(x,y)=(r\cos\theta,r\sin\theta)$. Then

$$ \lim_{(x,y)\to(0,0)}{f(x,y)}=\lim_{r\to0}{\frac{r^2\cos\theta\sin\theta+r^2\sin^2\theta}{r^4\cos^4\theta+r^2\sin^2\theta}}=\lim_{r\to0}{\frac{\cos\theta\sin\theta+\sin^2\theta}{r^2\cos^4\theta+\sin^2\theta}}=\frac{\cos\theta\sin\theta+\sin^2\theta}{\sin^2\theta}=1+\cot\theta, $$

so, because the limit depends on the value of $\theta$, then the limit does not exist, hence $f(x,y)$ is not continuous at $(0,0)$.

Is that correct? Can we use polar coordinates here?

Thanks!

$\endgroup$
  • 2
    $\begingroup$ "if $\lim_{(x,y)\to (0,0)}f(x,y)$ exists then it must be equal to $0$" - nope, the limit might exist, but differ than zero. It might be for example $3$. If it exists and equals $0$, then the function is continuous at the origin. Your computation is correct and so is your conclusion. If you want to avoid polar coordinates the paths $y=x$ and $y=x^2$ should work. $\endgroup$ – Galc127 Oct 25 '18 at 1:57
  • $\begingroup$ @Galc127 thank you. I do not think that I am wrong. Since the concept of "existence of a limit" implies the limit must be finite it can be any number, but since we want the limit to be zero (so it can be continuous), then [if it is a number] it must be equal to $0$. $\endgroup$ – manooooh Oct 25 '18 at 2:23
  • 1
    $\begingroup$ @Andrei I do not say only that. A statement before I said "I know that $f(0,0)=0$". We know that a function is continuous iff $f(x_0,y_0)=\lim_{(x,y)\to(x_0,y_0)}{f(x,y)}$. To make equality true, if $f(x_0,y_0)=f(0,0)=0$ then the other side must be equal to $0$ too. I do not see any forgotten term. $\endgroup$ – manooooh Oct 25 '18 at 3:40
2
$\begingroup$

Although your argument contains a grain of truth, it is not quite correct as it is written, since you wrote that the limit $\lim_{(x,y)\to (0,0)} f(x,y)$, which doesn't exist, is equal to the limit $\lim_{r \to 0} (\cdots)$, which does exist (for $\sin \theta \neq 0$) if you just view it as an ordinary single-variable limit which depends parametrically on a constant $\theta$; in fact, you just computed it yourself like that and wrote that it's equal to $1 + \cot \theta$.

The problem is that you want $\theta$ to be able to vary independently of $r$ as $r \to 0$, so you can't treat $\theta$ as a constant here. It should be viewed as an arbitrary function $\theta(r)$.

In fact, polar coordinates are mainly useful for proving that a limit exist, namely if you can write $f(x,y)$ as a bounded factor times another factor which depends only on $r$ (no $\theta$!) and which tends to zero as $r \to 0$ (really just a single-variable limit here!), then $f(x,y)\to 0$ as $(x,y)\to(0,0)$.

To show that a limit does not exist, you instead find two ways of approaching the point such that you get two different values. In your case, consider $f(t,0)$ and $f(0,t)$ as $t\to 0$, for example.

So actually I don't quite know how I would like to write the argument in a nice way if someone forced me to use polar coordinates in order to show that a limit does not exist! I would probably just write $f(x,y)$ in polar coordinates first, $$ f(r \cos\theta(r), r \sin\theta(r)) = \cdots, $$ (no “$\lim$” here) and then say that by making different choices of the function $\theta(r)$ (for example different constant functions!), you can make $f$ approach different values. And I would give examples of two such function $\theta(r)$ which give different limits for $f(r \cos\theta(r), r \sin\theta(r))$ as $r \to 0$.

$\endgroup$
  • $\begingroup$ Wow great answer, thank you! Yeah, what I done is a bit ugly from your reasons. The truth is that I only did iterated limits and by path, and now I wanted to "show" another approach (also I do not frequently use polar coordinates). So do you suggest not to go through this way? $\endgroup$ – manooooh Oct 25 '18 at 10:29
  • 1
    $\begingroup$ @manooooh: As I said, showing that a limit doesn't exist is usually easier directly in the original coordinates. $\endgroup$ – Hans Lundmark Oct 25 '18 at 16:05
  • $\begingroup$ Off-topic: are iterated limits a particular case of limits by path (parabolas, etc.)? $\endgroup$ – manooooh Oct 25 '18 at 16:56
  • 1
    $\begingroup$ @manooooh: Not really in general, although often the inner limit can be computed simply as follows: $\lim_{x \to 0} \left( \lim_{y \to 0} f(x,y) \right) = \lim_{x \to 0} f(x,0)$. And in that case, the iterated limit becomes the limit along a path, namely along the $x$-axis. $\endgroup$ – Hans Lundmark Oct 25 '18 at 17:33
  • 1
    $\begingroup$ Rather that both become meaningless. So the iterated limit doesn't exist, but the two-variable limit may very well exist, for example if $f(x,y)=x y \arctan(1/y)$ for $y \neq 0$. $\endgroup$ – Hans Lundmark Oct 26 '18 at 6:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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