Consider a problem of the form $f(x,y) = \frac{g(x,y)}{h(x,y)}$ for $x \neq y$ and $f(x,y) = a$ for $x=y$ such that $a$ is a fixed real number and $(x,y) \in \mathbb R^2$. And we need to show that this function is continuous for all values of $(x,y)$ in the domain.
Showing that $f(x,y)$ is continuous at all points $x \neq y$ is simple in the case where both the numerator and denominator are continuous and the denominator is not equal to $0$.
However, I'm not sure what the best approach is to show that the function is continuous for all $(x,y)$.
If there was an isolated point that needed to be considered (such as $(0,0)$), then I could simply treat that on its own. However, in this case, there are infinitely many points where $x=y$, and so I'm not quite sure what the best way to deal with this type of problem is.
I would be grateful for any input here.
This is not a complete answer.
If $f$ is translation invariant, i.e. $f(t+x,t+y)=f(x,y)$, then it suffices to check the continuity of $f$ at the origin.
If $f$ is general, then I guess we don't have a 'best' way to deal with its continuity along the line. The main reason is as follows.
For a translation invariant function, its behaviors around different points on the line are the same. We can call this property uniformly behaves (not a standard term. I just made it up for simplicity of statement). While for a general $f$, corresponding to general $g$ and $h$, the behaviors of $f$ around different points on the line can be quite different, i.e. differently behaves (again, just a made-up word corresponding to the one above). Thus, one should not expect that there is a 'simple' or 'best' way to identify all the continuous points of $f$ on the line.