Are functions still continuous even with removable discontinuities? If we define a continuous function as a function that is continuous at all points in its domain, then would a removable discontinuity not impact this?
For example consider the function $f(x) = x$. It is clearly continuous at every point in its domain, which is all real numbers.
But now let's have $f(x) = x \cdot \frac{x-2}{x-2}$ which means the function is undefined at $x=2$, but still equivalent to $f(x)=x$ everywhere else, so we exclude that point $x=2$ from the domain.
The domain is now something like $(-\infty, 2) \cup (2, \infty)$. Continuity at a point usually means the limits exist and they equal the function's value at that point. The number $2$ is not in the domain so we only have to consider numbers to the left and to the right of $2$, and no matter what number we pick, they will be defined and have limits equaling those values.
So would we still call such a function "continuous" despite having this $f(2)$ undefined?
 A: Yes, of course. The only thing that matters for continuity are those points at which the function is defined. What occurs outside the domain is irrelevant.
A: Yes. The function is continuous, because it is still continuous in every point of its (now smaller) domain $(-\infty,2)\cup(2, \infty)$.
I am aware that there are other "definitions" of continuity. The Wikipedia article is confusing in that sense, as it gives a number of definitions that are equivalent to each other, and some that are not:

The most common and restrictive definition is that a function is continuous if it is continuous at all real numbers.

Ugh! Come on, it means that a function not defined on the whole $\mathbb R$ cannot be continuous!
I am sure there are other sources of information that muddle the waters more than they make things clear...
Obviously, if you look at the (continuous) function $f(x)=x, f:\mathbb R\setminus\{2\}\to\mathbb R$ - its graph is not connected, has two components, but that is not what we call discontinuity. A discontinuity would be a point $x_0$ such that $\lim_{x\to x_0}f(x)\ne f(x_0)$, and to even check this condition you must be able to calculate $f(x_0)$, i.e. the function $f$ must be defined in $x_0$ (in our case, $x_0\ne 2$).
Thus, my conclusion for the OP is this:


*

*Take the universally accepted definition of continuity: the function is continuous if it is continuous at every point of its domain. Or any definition equivalent to that one. (I personally like the topological one, where the function is continuous if the preimage of any open subset of the codomain is an open subset of the domain.)

*Accept that the graph of a continuous function need not be a connected set (in, say, $\mathbb R^2$) - though, obviously, it will be if the domain of the function itself is connected.

*Accept that in maths' education there are other non-equivalent definitions of continuity floating around; fight them with arguments (even if it seems like attacking windmills).
A: Would like to add to the other answers:
Agreed that a function which is continuous on the domain that it's defined is "continuous". However, going by the title and generalizing a bit depending on your definition of removable discontinuities and then depending on the function, that function may still not be continuous. In your question description you talk about if f(2) is undefined, but that may not be the ONLY case where you have removable discontinuities. 
A definition may allow a function with removable discontinuities to be defined at the discontinuous points. For example, f(x) = x for all x in R except x = 2, for which f(x) = 1. This function is truly discontinuous, and the removable discontinuity is truly a discontinuity. This is similar to how one might use/make sense of the term "infinite" discontinuity", for example f(x) = 1/x for non-zero x, and f(x) = 0 for x = 0. In this case, f is a discontinuous function. 
Of course, the terms probably originated because people were talking about the whole Real number line instead of the domains in which the functions are defined, and for better or for worse (probably for worse) the naming stuck.   
A: It is correct but note that in some context a function is defined to be continuos if its domain is an interval, and it is continuous at every point of that interval.
Here you can find a good source from MIT for a full classification Continuity and Discontinuity.
