Given $g(x,y) = \begin{cases} \frac{x^2-y^2}{x-y}, & \text{if $x \neq y$}\\[2ex] 3x, & \text{if $x=y$} \end{cases}$
Determine all points at which $g$ is continuous.
In order to establish continuity we must determine the limit of $g(x,y)$ as $(x,y)→0$. From what I have found so far,
$\lim\limits_{(x,y)\to (0,0)}\frac{x^2-y^2}{x-y}$ $=$ $\lim\limits_{(x,y)\to (0,0)} (x+y)$ $=0$
$\lim\limits_{(x,y)\to (0,0)}3x = 3(0)=0$
Hence, $\lim\limits_{(x,y)\to (0,0)}g(x,y) = 0$. However, I am unsure how to proceed from here as I can only safely assume $(0,0)$ is a point of continuity, but how do I go about finding the other points of continuity?
$f$ isn't continuous at any point on the line $y=x$ other than the origin, by the limit calculation:
$\lim_{(x,y)\to (x,x)}g(x,y)=2x$.
Because by the limit point definition of continuity, we need the limit to be $g(x,x)=3x$. This is only true at the origin.