Question

let f(x,y)= (x+y)^2/ x-y
we claim that it cannot be extended over (0,0), is that true?

Answer 1

What do you mean by 'cannot be extended over (0,0)'

the limit might exist, so you could redefine f(0,0)

Answer 2

f(0,0) is undefined but if the claim true i need help finding a curve or direction from our domain which is |x|>|y|

Answer 3

what can we choose y from the domain to show that the claim is true?

Answer 4

Sorry I'm having trouble understanding the language
of the claim.

Answer 5

Are you trying to extend the function so it is continuous?

Answer 6

yes

Answer 7

for example let y= mx where 0<=m<1 so the lim will go to 0

Answer 8

what else can we choose y to be?

Answer 9

Ok, let's try that. y = x

Answer 10

but our domain is |x|>|y|, is that work?

Answer 11

if y = x we have
(2y)^2 / (y - y ) = 2y^2 / 0 , which is a nonremovable discontinuity

Answer 12

I don't think it should make a difference that |x| > |y| .

Answer 13

so that is not continuous so it cannot extend over (0,0)?

Answer 14

what if we let y= x/2 , does it work?

Answer 15

Sure. Just let f(x,y)=(0,0) at (0,0).

Answer 16

I did that but my professor asked for a curve or direction from our domain which is |x|>|y| that satisfies the claim

Answer 17

sorry. i absolutely misread your question. have you considered \[\gamma\left(t\right)=\left(x,x\cos x\right)?\]
It's a smooth curve such that\[\lim_{t\to0^-}f\left(\gamma\left(t\right)\right)=-\infty\qquad\text{and}\qquad\lim_{t\to0^+}f\left(\gamma\left(t\right)\right)=+\infty.\]

Answer 18

This shows that \(f\) cannot be extended at \((0,0)\)