# Continuity of a Function of Two Variables

1. Nov 8, 2004

Question:
Is the function f(x,y) = (x^2 - y^2)/(x-y) continuous at (1,1) if we set f(1,1) = 0? Why or why not?

So far, I've just plugged 1 in for x and y and found the limit to equal 0. I guess that means that the limit is not continuous at (1,1)? And what do they mean by set f(1,1) = 0?

2. Nov 8, 2004

### Galileo

The function f(x,y) = (x^2 - y^2)/(x-y) is not defined at (1,1), since:
(1-1)/(1-1)=0/0.
So setting f(1,1)=0, means defining a value for the function at (1,1).

By the way. The limit is not equal to 0.
Note that : (a+b)(a-b)=a^2-b^2.

3. Nov 8, 2004

### ReyChiquito

$$\lim_{(x,y)\rightarrow(1,1)}f(x,y)=\lim_{(x,y)\rightarrow(1,1)}\frac{x^2-y^2}{x-y}=\lim_{(x,y)\rightarrow(1,1)}\frac{(x+y)(x-y)}{x-y}=\lim_{(x,y)\rightarrow(1,1)}x+y=2$$

Last edited: Nov 8, 2004