calculation – How can a function with a hole (removable discontinuity) be equivalent to a function without a hole?

I have done some research and hope that someone will be able to control me. My question was:

Suppose I have the function $$f (x) = frac {(x-3) (x + 2)} {(x-3)}$$, so he has removable discontinuity to $$x = 3$$. We remove this discontinuity with algebra: $$f (x) = frac {(x-3) (x + 2)} {(x-3)} = (x + 2)$$. BUT, the graph of the first function has a hole to $$x = 3$$, 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?

Because the original function is undefined at the point $$x = 3$$we have to limit the domain to $$mathbb {R} setminus 3$$. And when we manipulate this function with algebra, the final result, $$f (x) = (x + 2)$$ always use this restricted domain. So, even if the function $$f (x) = (x + 2)$$ would not have a hole if the domain were all $$mathbb {R}$$, we're sort of "imposing" a hole in $$x = 3$$ by continuing to throw this 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 are working with a function that is everywhere continuous, which helps us to easily find the limit. But the reality is that the function $$f (x) = (x +2)$$ is in fact NOT continuous everywhere when we restrict the domain by rejecting point 3. Or am I pushing things too far?