Continuity of $\begin{cases}(xy+y^2)/(x^4+y^2)&\text{if }(x,y)\neq(0,0),\\0&\text{if }(x,y)=(0,0)\end{cases}$ at origin using polar coordinates

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!

• "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 1:57
• @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 2:23
• @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 3:40

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 $$r \to 0$$.

• 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
• @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
• @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
• 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 6:36