0

I have: \begin{equation} f_1(x,y)=\frac{xy}{(x^2+y^2)^{\frac{1}{2}}} \end{equation}

Is this function continuous at the origin, still it does not exist there?

Thanks

Luthier415Hz
  • 2,739
  • 6
  • 22
  • 3
    In polar coordinates the function is $r\cos\theta\sin\theta$, which does have the value $0$ at the origin (or if you want to be pedantic, the singularity is removable). – Parcly Taxel Oct 13 '21 at 12:29
  • This requires a transformation. If f is to be defined for continuity on R, should transformations be included from Cartesian systems? – Luthier415Hz Oct 13 '21 at 12:33
  • You would transform the origin too to radial coordinates. – Luthier415Hz Oct 13 '21 at 12:34
  • Your computations don't make much sense. Just from looking along the axes (is that what you're doing? – it's not clear what you mean by “$(x,y) \to (0,x)$”, for example), you can't conclude that the multivariable limit exists. And your “multivariable l'Hospital's rule” is also wrong; the expression “$f'(x,y)/g'(x,y)$” is meaningless! – Hans Lundmark Oct 13 '21 at 13:13
  • @Hans Lundmark https://math.stackexchange.com/questions/1203220/when-can-lhospital-rule-be-used-on-multivariable-limits – Luthier415Hz Oct 13 '21 at 14:46
  • @vqngs: Yes? What do you want to say with that? – Hans Lundmark Oct 13 '21 at 17:24

2 Answers2

4

Functions can only be continuous at points in their domain. $(0,0)$ is not in the domain of $f_1$, so speaking of continuity at $(0,0)$ is nonsensical.

Note that, as others have pointed out, there exists an extension of $f_1$ which is in fact continuous at the origin. That function is

$$F:\mathbb R^2\to\mathbb R$$ $$F(x,y)=\begin{cases}f_1(x,y); & (x,y)\neq (0,0)\\ 0;&(x,y)=(0,0)\end{cases}.$$

The function $F$ is equal to $f_1$ on the entire domain of $f_1$, but the function $F$ also contains $0$ in its domain, and it is continous at $0$.

Note also that $F$ is the only function on $\mathbb R^2$ that both extends $f_1$ (i.e., returns the same value as $f_1$ on all points in the domain of $f_1$) and is continuous.

5xum
  • 123,496
  • 6
  • 128
  • 204
  • Which theorem states this? – Luthier415Hz Oct 13 '21 at 12:29
  • 5
    @vqngs The definition of continuity. – 5xum Oct 13 '21 at 12:29
  • Thanks for that. I will look this theorem up. – Luthier415Hz Oct 13 '21 at 12:32
  • 3
    @vqngs It is not a theorem, it is a definition. – 5xum Oct 13 '21 at 12:32
  • Thanks, I understand. But what about @Parcly Taxel comment? – Luthier415Hz Oct 13 '21 at 12:33
  • The limit can exist as you approach a point outside the domain of definition. That means you can extend the domain but it does not automatically make that point a part of the domain. – Clayton Oct 13 '21 at 12:34
  • That comment says that you can extend the domain so that the function is defined at the origin. And that you can do it in such a way that the new function is continuous at the origin. – Arthur Oct 13 '21 at 12:35
  • To be clear: you can create a new function that agrees with the old function at all points in the domain of the old function. The new function, though, could have a strictly larger domain (such as including the origin in the OP), and the new function can be continuous there because it’s part of the domain. – Clayton Oct 13 '21 at 12:36
  • The domain is R, R includes the origin, but at origin f does not exist. But a limit exist as R approaches the origin, so this is discontinuity of the first kind, since there is an asymptote at 0,0 – Luthier415Hz Oct 13 '21 at 12:37
  • 2
    No. If the function doesn't exist / doesn't have a value / isn't defined at a point, then that point is not in the domain of the function. That's what "domain" means. – Arthur Oct 13 '21 at 12:39
  • 1
    @5xum, the definition of continuity is that , for two metric spaces, X and Y, and f maps E into Y, where E subset of Y at p, for every epsilon>0, then there exists a delta>0 such that: d_Y(f(x),f(p))<\epsilon . So at p=0, d_Y(f(x,y), f(0,0))<\epsilon , but since f(0,0) does not exist, then d_Y((f(x,y), f(0,0)) does not exists, and there is no epsilon that can quantify continuity. You are right. – Luthier415Hz Oct 13 '21 at 12:42
  • @vqngs Exactly. But, as others have pointed out, there exists an extension of $f_1$ which is in fact continuous at the origin. That function is

    $$F:\mathbb R^2\to\mathbb R$$ $$F(x,y)=\begin{cases}f_1(x,y); & (x,y)\neq (0,0)\ 0;&(x,y)=(0,0)\end{cases}$$

     – 5xum Oct 13 '21 at 12:46
  • @vqngs More fundamentally, note on your definition that $p$ is explicitly specified to be in the domain of $f$. – Arthur Oct 13 '21 at 12:47
  • @vqngs The function $F$ is equal to $f_1$ on the entire domain of $f_1$, but the function $F$ also contains $0$ in its domain, and it is continous at $0$. – 5xum Oct 13 '21 at 12:47
  • @5xum thanks. cannot change that definition anymore, it is locked Arthur. – Luthier415Hz Oct 13 '21 at 12:49
1

As others pointed out, the proper question is "Can $f$ be extended to $(0,0)$ as a continuous function?". The answer is yes because, since $\lim_{(x,y)\to(0,0)} f(x,y)= 0$, defining $f(0,0)=0$ automatically yields a continuous function.

You can see this simply by noting that $$ \left| \frac{xy}{\sqrt{x^2+y^2}}-0 \right|\leq \dfrac{|x||y|}{\sqrt{x^2+y^2}}\leq \dfrac{\sqrt{x^2+y^2} \sqrt{x^2+y^2}}{\sqrt{x^2+y^2}} = \sqrt{x^2+y^2} \to 0. $$

I used the inequalities $$ |x| = \sqrt{x^2}\leq \sqrt{x^2+y^2}$$ $$ |y| = \sqrt{y^2}\leq \sqrt{x^2+y^2}$$

PierreCarre
  • 20,974
  • 1
  • 18
  • 34