Can we talk about the continuity/discontinuity of a function at a point which is not in its domain? Let us say that I have a function $ f(x)=\tan(x)$ we say that this function is continuous in its domain.
If I have a simple function like $$ f(x)=\frac{1}{(x-1)(x-2)} $$
Can we really talk about its continuity/discontinuity at $x=1$ or at $x=2$.
From what I know we can't since it is not in its domain.
But doesn't it make every function of the form $$ f(x)=\frac{1}{g(x)}$$ continuous in its domain. Where $g(x)$ is any polynomial and $g(x)=0$ at n points (lets say).
 A: As Peter Dombrowski put it, roughly translated: 
“The function $1/x \colon \mathbb R^* \to \mathbb R$ is not defined in $0\in\mathbb R$.  Hence for us the following group of symbols and words: 
‘$1/x \colon\mathbb R^*\to\mathbb R$ is not continuous in $0$.’
is no statement, i.e., neither true nor false.”
A: A function by definition must be defined for all points in the domain. So formally speaking a function like $\tan(x)$ doesn't even know what $\pi/2$ is (other than the codomain maybe). So no, it does not make sense to talk about continuity at points outside the domain. 
For example is the function $f\colon \Bbb{R} \to \Bbb{R}$, $x\mapsto x$ continuous at the point $x=\text{New York}$? It is just as meaningless to talk about $\tan(x)$ being continuous at $\pi/2$ as it is to talk about it being continuous at $\text{New York}$.
Another way to see this is the definition. A function is continuous at a point $a$ iff:
$$\lim\limits_{x\to a} f(x) = f(a)$$
Well if the right side doesn't exist, this clearly can't be true.
A: What you can ask is whether that function can be continuously extended also at $x=1,2$, i.e. whether given $f$ defined on, say, $\mathbb R\setminus\{ x_0 \}$ there exists a function $g$ defined and continuous on the whole $\mathbb R$ such that its restriction $g\|_{\mathbb R\setminus\{ x_0 \}}$ is equal to $f$. This may happen and it is called removable singularity, or removable discontinuity, though, as you say, it is not properly a point in which the function is discontinuous, since it is not defined there.
Rational function singularities, i.e. of the form $\frac{P(x)}{Q(x)}$ where $P$ and $Q$ are polynomials and $Q(x_0)=0$ are removable only if $P(x_0)=0$ as well and the multipliciy of this root in $P$ is equal or higher to the multiplicity of the root in $Q$ (in other words if you can simplify a common factor of the form $(x-x_0)^n$ in both polynomials). In the case $\frac 1{g(x)}$ with $g$ polynomial you can say that the function is always continuous in its domain but singularities are not removable.
A: From p. 113 of Calculus by Spivak:

The function $f(x) = \sin 1/x$ is not continuous at $0$, because it is
  not even defined at $0$, and the same is true of the function $g(x) = x\sin 1/x$.

