Removable discontinuity question A quick and easy question on terminology:
If we have a function $$f(x) = \frac{x(x-2)}{(x-2)}$$ then clearly the function is not defined at $x = 2$.
If we cancel out "$(x-2)$" then the function is defined everywhere (it is just the identity function).
Does the definition of function not require that we consider these to be two different functions?
 A: The most usual definition of a function, taught ever since people decided to reduce mathematics to set theory (about a century ago), regards these as two different functions.  And by that definition, Dirac's delta function and its derivatives and related things are not functions at all.  I think occasionally someone urges other approaches than the usual reduction to set theory.
A: Yes. That is correct. The functions are equal at all values of $x$ except $x=2$. The graph of the first function is the line $y=x$ with a hole at the point $(2,2)$.
A: NOTE: The concept of "removable discontinuities" in elementary calculus is a weird one.  The function $f$ is not really discontinuous over its domain.  This is because $2$ is not really in its domain.
The function $f$ as you've defined it has the signature $f: \Bbb R \setminus \{2\} \to \Bbb R$.  This means that the function $f$ is defined everywhere in the real numbers $\Bbb R$ except at $2$.  If this signature (or just the domain) is specified first, then we could just as well define the mapping $f$ by $f(x) = x$.  Because $x=\dfrac {x(x-2)}{x-2}$ for all values $x\ne 2$.
All you are doing when you "cancel out" the $(x-2)$'s is defining a new function $f_{new}$ which is an extension of $f$ to the topological closure of its domain.  Basically the topological closure of a set $X$ has every element of the set $X$ AND every limit point of the set $X$.  In this case $2$ is a limit point of the set $\Bbb R \setminus \{2\}$, so the closure of the set $\Bbb R \setminus \{2\}$ is just the set $\Bbb R$.  Then the way we define the value of $f_{new}(2)$ is by making the function continuous at that point (not always doable, but in this case it is).  So $f_{new}(2) = \lim_{x\to 2} f(x) = 2$.
So this function $f_{new}$ takes the same value as $f$ at every point in the domain of $f$ and additionally takes the value $2$ at $x=2$.  So in this case, $f_{new}$ has the signature $f_{new}: \Bbb R \to \Bbb R$ and is defined by $f_{new}(x)=x$.  Are $f$ and $f_{new}$ the same function, then, if they can both be defined by $f(x)=x$?  No.  They are different because they have different signatures: $f$ is NOT defined at $x=2$, but $f_{new}$ is.
A: A function $f$ is a relation $f\subseteq D\times C$, such that 
$$((x,y),(x',y')\in f\wedge x=x')\implies y=y',$$
where $D$ is the domain and $C$ is the co-domain of the function. The domain of the identity $id$ of $\mathbb R$ is whole $\mathbb R$ while the domain of $f$ is 
$\mathbb R \setminus\{2\}$, so $id\ne f$.
A: Well, we know that the identity function ($f(x)=x$) is the product of the limit 
$$\lim_{x\rightarrow2} f(x) = \frac{x(x-2)}{x-2}.$$ Thus making them two equal functions except at $x=2$ due to the limit on the function from $x\rightarrow2$. There will just be a hole at $x=2.$ So yes, you are correct. They are different.
A: Two functions are considered equal if they have the same domain (so your function is not the same as the identity function, as others have pointed out), the same codomain (set where the values of the function live), and agree at each point of their common domain. This provides the interesting  situation that if $X$ is a proper subset of a set $Y$, then the identity function on $X$ is considered a different function from the inclusion map of $X$ into $Y$, because the codomains are not the same for each function.
