function identity: does $\frac{x^2-4}{x-2} = x+2$ Can you tell me how to resolve the (apparent) paradox that the function:
$$f(x) = \frac{x^2-4}{x-2}$$
is identical to the function:
$$g(x) = x+2$$
because by factoring the numerator:
$$f(x) = \frac{(x+2)(x-2)}{x-2} = x+2$$
But $f(x)$ can't be identical to $g(x)$ since the former is undefined at $x = 2$ and the latter is not ?
Thanks,
Mike
 A: By defintion two functions are said to be identical if they have the same domain  and the same image so $f\neq g$ but we may say that $f$ is extended at $2$ by continuity on the function $g$.
A: This is largely a difficulty of notation. Your definition doesn't explicitly state what to do in the case $x = 2$. Some people would take this omission to mean $f$ is undefined there; others would take it to mean $f$ is the natural thing for it to be, the only value that makes it continuous at that point. Usually the latter interpretation is harmless, in which case the two functions really are equal.
Supposing we don't like this imprecision, and we demand that $f$ is undefined at $2$ so has a strictly smaller domain than $g$. In this case, they are certainly not equal, but you have to be careful. I'd say that $f \not= g$, but it's wrong to say that $f(x) \not= g(x)$, so in particular it's wrong to say that $\frac{x^2 - 4}{x-2} \not= x + 2$: either the equation is true or meaningless, depending on what you think $x$ is.
A: They are not identical. The first one has what we call a removable singularity at $x=2$. There are times when we don't care about the difference, though, and we just consider a removable singularity to be filled in.
A: It's true for all $x$ except $2$, keeping in mind the discontinuity at $x=2$.
A: It may be helpful to think of a function of one variable like this as a set of ordered pairs. Here, $(2,4)\in g$, but $(2,4)\notin f$. So, despite the fact that all other elements of $g$ are also elements of $f$, $f$ and $g$ are not the same functions.
