# Continuous extension(2)

1. Aug 11, 2010

### Alexx1

How can the function

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

be defined on the line y=x so that we get a continuous function?

Is this correct?: If x=y --> f=0

2. Aug 11, 2010

### HallsofIvy

No. Did you have any reason at all for thinking that?

For $x- y\ne 0$
[tex]\frac{x^3- y^3}{x- y}= x^2+ xy+ y^2[/itex]

As long as x is NOT equal to y, that function is the same as $x^2+ xy+ y^2$. The function will be continuous on the line the line x= y, if we define $f(x, y)= f(x, x)= x^2+ xy+ y^2= x^2+ x^2+ x^2= 3x^2$, not "0".

3. Aug 11, 2010

### Alexx1

Thank you very much!