I've done some research, and I'm hoping someone can check me. My question was this:
Assume I have the function f(x) = \frac{(x-3)(x+2)}{(x-3)}, so it has removable discontinuity at x = 3. We remove that discontinuity with algebra: f(x) = \frac{(x-3)(x+2)}{(x-3)} = (x+2). BUT, the graph of the first function has a hole at x = 3, and the graph of the second function is continuous everywhere. How can they be "equal" if one has a hole and the other does not?
I think that this is the answer:
Because the original function is undefined at the point x = 3, we have to restrict the domain to \mathbb{R} \setminus 3. And when we manipulate that function with algebra, the final result, f(x) = (x + 2) is still using this restricted domain. So even though the function f(x) = (x+2) would not have a hole if the domain were all of \mathbb{R}, we are sort of "imposing" a hole at x = 3 by continuing to throw that point out of the domain.
And then just to close the loop: Removing the removable discontinuity is useful because it allows us to "pretend" that we're working with a function that is everywhere continuous, which helps us easily find the limit. But the reality is that the function f(x) = (x +2) is actually NOT continuous everywhere when we restrict the domain by throwing out the point 3. Or am I now taking things too far?
Thanks in advance!
EDIT: For anyone coming across this in the future, in addition to the excellent answers below, I also found this other question about the continuity of functions with removable discontinuities helpful.
