Today I saw an example of finding the domain of a multivariable function. The function in question was this:
$$f(x,y)=\frac{\sqrt{y-x^2}}{1-x^2}$$
The solution:
For the function, we only want real values so $y$ must be larger than or equal to $x^2$. Also, we can't divide by zero, so $x$ cannot be 1.
They say $x$ cannot be one, since this would be division by 0. But what if we took the point $(1,1)$? Then we would have $\frac{0}{0}$, which is indeterminate. So what's to say we couldn't find that the function is defined at $(1,1)$ after all, by taking some kind of limit to the point? I haven't actually tried to, but that's not the point (no pun intended). Why is saying that the denominator is zero for $x=1$ sufficient to prove the function is not defined at that point?
Because $y=1\implies x=\pm 1$ (in the minimum case) which is in turn ruled out by the $x\neq \pm 1$ condition.