1

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!

manooooh
  • 2,299
  • 2
    "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. – Galc127 Oct 25 '18 at 01:57
  • @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 $0$. – manooooh Oct 25 '18 at 02:23
  • 1
    @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. – manooooh Oct 25 '18 at 03:40

1 Answers1

4

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 $r \to 0$.

Hans Lundmark
  • 55,038
  • 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? – manooooh Oct 25 '18 at 10:29
  • 1
    @manooooh: As I said, showing that a limit doesn't exist is usually easier directly in the original coordinates. – Hans Lundmark Oct 25 '18 at 16:05
  • Off-topic: are iterated limits a particular case of limits by path (parabolas, etc.)? – manooooh Oct 25 '18 at 16:56
  • 1
    @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. – Hans Lundmark Oct 25 '18 at 17:33
  • Why not in general? Counterexample? – manooooh Oct 25 '18 at 18:07
  • @manooooh: Well, $\lim_{y \to 0} f(x,y)$ doesn't have to equal $f(x,0)$. For example, if $f(x,y)$ is only defined for $y \neq 0$. – Hans Lundmark Oct 25 '18 at 19:12
  • So are you saying in that case $\lim_{x \to 0} \left( \lim_{y \to 0} f(x,y) \right) \neq\lim_{x\to0}f(x,0)$? – manooooh Oct 25 '18 at 21:57
  • 1
    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 $y \neq 0$. – Hans Lundmark Oct 26 '18 at 06:36