Answer:
It cannot be extended.
Step-by-step explanation:
Consider the function [tex]f(x,y) = \frac{x^2-y^2}{x^2+y^2}[/tex]. To extend this functions so it is continous at (0,0) we must define [tex] f(0,0) = \lim_{(x,y)\to(0,0)\frac{x^2-y^2}{x^2+y^2}[/tex]. However, this implies that the limit exists. So, we should find if the limit exists or not.
In this case, consider the case in which y =0. When y=0 then
[tex]\lim_{(x,y)\to(0,0) \frac{x^2-0^2}{x^2+0^2} = \lim_{x\to 0}\frac{x^2}{x^2}= 1[/tex]
But, when x=0, we get
[tex]\lim_{(x,y)\to(0,0) \frac{0^2-y^2}{0^2+y^2} = \lim_{y\to 0}\frac{-y^2}{y^2}=-1[/tex].
So, since the limit depends on how we approach to the point (0,0) the limit does not exist. So we can't extend f(x,y) so it is continous.
