How can f(x, y) be defined on y=x for a continuous function?

[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

 

[HallsofIvy]

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

For x- y\ne 0
\frac{x^3- y^3}{x- y}= x^2+ xy+ y^2[/itex]<br /> <br /> 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 &quot;0&quot;.

 

[Alexx1]

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

For x- y\ne 0
\frac{x^3- y^3}{x- y}= x^2+ xy+ y^2[/itex]<br /> <br /> 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 &quot;0&quot;.

<br /> <br /> Thank you very much!