Continuity at the origin

fskilnik

$f\left( {x,y} \right) = \left\{ \begin{array}{l} \frac{{x^2 \,y}}{{x^3 - y^2 }}\,\,\,\,if\,\,\,x^3 \ne y^2 \\ \,\,\,\,\...$
Is f continuous at (0,0) ?

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The Captain · Riemann Hypothesis Offline Joined: 08 Feb 2005 Posts: 388

No!
A classical one. I'll introduce you to a nice way .Let y=1*x,then the result will be clear.
Check y=f(x)*x,where lim f(x)=1 as x goes to 0.
For instance f(x)=e^x ,(1-x),... .The latter will do the trick for us.

**Posted:** Wed Jul 20, 2005 1:43 am

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fskilnik

Thank you for your reply, "Captain".

The idea is very nice but when I tried y=x(1-x) I got $\mathop {\lim }\limits_{x\, \to \,0} \,\,f\left( {x\,,\,\,x\left( {1 - x} \right)} \right) = 0 = f\left( {0,0} \right)$ Therefore, if I am not wrong, it seems to me that the problem is still "open" !

P.S.: I did think about some "intrinsic relationship" between x and y to find "problematic sequences" but... I must confess it did not last long till I started trying my (impossible?) mission of proving the continuity at this point!

**Posted:** Wed Jul 20, 2005 10:49 am

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The Captain · Riemann Hypothesis Offline Joined: 08 Feb 2005 Posts: 388

I just tried to give you a hint ,and I thought it would help.
Now consider this one $y=(1-x)x^\frac{3}{2}$ . This will lead us to the desired result.

**Posted:** Sat Sep 17, 2005 3:21 am

Jimmy · Yang-Mills Theory Offline Joined: 03 Oct 2004 Posts: 920

Solution was wrong.

fskilnik

Very nice solutions indeed. Smile

Thank you both, Captain and Jimmy.

**Posted:** Sun Sep 18, 2005 12:25 pm

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The Captain · Riemann Hypothesis Offline Joined: 08 Feb 2005 Posts: 388

Dear Jimmy.You're not suggesting the right thing,I'm afraid.
First of all the statement you have mentioned, Does Not hold in general.
(If you wanna see why ,you can refer to articles about "Uniform Convergence".)
Another point is that $x^2y=r^3\cos^2\theta\sin\theta$ !! (A typo).

**Posted:** Mon Sep 19, 2005 1:09 am

fskilnik

Dear Captain,

Although the "r" missing in Jimmy´s calculations "collapses" his solution (perhaps it could be fixed with some $\alpha$ in $r^{\,\alpha } \cos \theta$ , "analogously" to your solution ?!) , I do not understand why the existence or not of uniform convergence "breaks" his reasoning (it certainly "disturbs" it but... does it break?) . Let me explain you why (and please reply about it):

IF his trigonometric ratio were right, let us say that for some pre-fixed $\theta _1 \ne \theta _2$ we got two sequences $\left( {r\cos \theta _{\,{\rm{i}}} \,\,,\,\,r\,sen\,\theta _{\,{\rm{i}}} } \right)\mathop \to \limits_{r\,\, \to \,\,0} \lef...$ Although the limits themselves $\left( {f\left( {r\cos \theta _{\,{\rm{i}}} \,\,,\,\,r\,sen\,\theta _{\,{\rm{i}}} } \right)} \right)_{\,r\,\, \to \,\,0} \, ...$ could NOT BE $\, \, \, \, \, - \frac{{\cos ^2 \theta _{\,{\rm{i}}} \,}}{{sen\,\theta _{\,{\rm{i}}} }}$ (yes, because of non-uniform convergence!) , couldn´t we argue that :

1) If one of them (at least) does not exist , then the discontinuity of f at the origin follows ( ${\theta _{\,{\rm{i}}} }$ is constant!)

2) If both of them exist, then they are different and the same conclusion applies?

Well... It seems to me that my initial question is much more interesting than I could suspect! Smile

Thank you both, Captain and Jimmy. I am looking forward to "hearing" from you!

Regards,
fskilnik

**Posted:** Mon Sep 19, 2005 9:18 am

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"Do not worry about your difficulties in Mathematics. I can assure you that mine are still greater." (A.Einstein)

Jimmy · Yang-Mills Theory Offline Joined: 03 Oct 2004 Posts: 920

**Posted:** Tue Sep 20, 2005 12:17 pm

The Captain · Riemann Hypothesis Offline Joined: 08 Feb 2005 Posts: 388

I start with Jimmy's question.You see, $\theta$ is not a constant; it's a variable depending on $x$ and $y$ .
And you want the limit not just as $r$ goes to zero, but on small disks around the pole,because you are
taking a two-dimensional limit.
You said you got $0$ .The reason is that a limit of $L$ as $x$ and $y$ approach zero means that $f(x,y)$ is near $L$ whenever $x$ and $y$ are BOTH near zero. Well, $x$ and $y$ are both near zero exactly when $r$ is near zero, but $\theta$ can have any value.
Now fskilnik question.
If the limit on the right exists for every $\theta$ , that is not enough to say that the limit on the left exists.
If the limit on the right exists for every theta and is independent of $\theta$ (i.e. it's always the same limit), then that is still not enough to say that the limit on the left exists. But if the limit on the right exists uniformly for all $\theta$ , then it is the same as the limit on the left. That is, the $\delta$ value for a given $\epsilon$ can't
depend on $\theta$ .

**Posted:** Wed Sep 21, 2005 2:41 am

_________________
"If calculation is the first thing that enters your mind when thinking of mathematics, I can assure you that your knowledge of math does not exceed simple addition and multiplication." Ehssan Khanmohammadi

fskilnik

Dear Captain,

Thank you for your response. I agree with most of it, but I am afraid you missed my point. Please let me explain it a bit more and please answer it whenever you have time.

First of all, let me discuss Jimmy´s question and your argument!

Jimmy: you may use polar coordinates "whenever" you want because all sequences
$\left( {x,y} \right) \in R^2$ are related to some sequence(s) $\left( {r,\theta } \right) \in \,\,R_ + \times \,\,\left[ {0,2\pi } \right[$ and reciprocally.

In other words, you are able to converge to the origin using rectangular coordinates or polar ones and topologically speaking there is no "preferences". We use one or the other (or others) according to the calculations involved!

The "problem" with your solution is that you got 0 (that´s right) and $f\left( {0,0} \right) = 0$ (by definition!) , therefore you could NOT prove discontinuity. The Captain proved it (discontinuity) , because the limit involved in his suggestion did not exist and SHOULD be zero to make continuity POSSIBLE. (This will be clearer in my comments to the Captain.)

I thought you got the discontinuity correctly proved, because I did not check your calculations till The Captain mentioned the "r" missing but, as I will explain now, if that ratio (you got) was right, you would have done it.

Captain:

Your phrase

**Posted:** Wed Sep 21, 2005 6:45 am

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"Do not worry about your difficulties in Mathematics. I can assure you that mine are still greater." (A.Einstein)

Jimmy · Yang-Mills Theory Offline Joined: 03 Oct 2004 Posts: 920

Thank you for your explanations. My math teacher just told us that if the limit as $r\to0$ cnverged to a constant, then that was the limit, and I never really questioned it, obviously I should have. Smile

And just to clarify, I was right to say that a limit depending on $\theta$ did not converge (which interestingly my math teacher never mentioned), right?

Thanks again.

**Posted:** Thu Sep 22, 2005 9:01 pm

fskilnik

Hello again, Jimmy!

The answer to your question is "right!" (I mean "yes") , in the sense that "if the limit exists, it is necessarily unique". (Limits don´t "converge/diverge", they exist or not!)

This problem is closely related to the following

Theorem: A function $\, \, f: R^{\,k} \to R^{\,m} \,\,\,\,\,\left( {for\,\,us: \,\,\,k = 2\,,\,\,m = 1} \right) \, \,$ is continuous at a certain given point $\, a \in R^{\,k} \, \,$ if and only if $\mathop {\lim }\limits_{n\,\, \to \,\,\infty } f\left( {x_n } \right)\,\, \to f\left( a \right) \, \,$ , for all sequences $\, \left\langle {x\,_n } \right\rangle _{n\,\, \ge \,\,1} \, \, \,$ in $\, \, R^{\,k} \, \, \, \,$ such that $\, \, \mathop {\lim }\limits_{n\,\, \to \,\,\infty } x\,_n \,\, = \,\,a \, \,$

Anyway, The Captain´s arguments are very important/instructive. You will learn a lot studying the "uniform converge" theme he suggested! (Before that, start with "pointwise-convergence" topics that you can find in good analysis books!)

All the best,
fskilnik.

**Posted:** Fri Sep 23, 2005 8:05 am

_________________
"Do not worry about your difficulties in Mathematics. I can assure you that mine are still greater." (A.Einstein)