Continuous Extension of Function at Origin: Can f=1?

[Alexx1]

How can the function

f: ℝ² → ℝ : (x,y) |--> $${{x^2+y^2-(x^3y^3)}\over{x^2+y^2}}$$ if (x,y) ≠ (0,0)

be defined in the origin so that we get a continuous function?When I take 'x=y' (so (y,y)) and 'y=x' (so (x,x)) I get:

$${{2-y^4}\over{2}}$$

and

$${{2-x^4}\over{2}}$$

So for the first one I get '1' when y=0
and for the second one I get '1' when x=0

So can I say that if (x,y)=0 --> f=1 ??

 

[Petr Mugver]

Yes, you can extend your function to a continuous one if you define f(0,0)=1.

 

[Alexx1]

Yes, you can extend your function to a continuous one if you define f(0,0)=1.

Thx!

 

[arildno]

In order to make this general, you should switch to polar coordinates.

Then, we can transforme the fraction to:
$$\frac{r^{2}-r^{6}\sin^{3}\theta\cos^{3}\theta}{r^{2}}=1-r^{4}\sin^{3}\theta\cos^{3}\theta, r\neq{0}$$
clearly, this is readily extendable for r=0.

 

[Alexx1]

In order to make this general, you should switch to polar coordinates.

Then, we can transforme the fraction to:
$$\frac{r^{2}-r^{6}\sin^{3}\theta\cos^{3}\theta}{r^{2}}=1-r^{4}\sin^{3}\theta\cos^{3}\theta, r\neq{0}$$
clearly, this is readily extendable for r=0.

Thank you very much!
And how about this one?

https://www.physicsforums.com/showthread.php?t=421612