# 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).

• – Hans Lundmark Oct 16 '15 at 7:35

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.”

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 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.

• This is a bit nitpicky, but your definition is wrong: it doesn't work for isolated points of the domain, where the limit does not exist, but the function is continuous nevertheless (this follows immediately from the correct definition). Other than that, I agree with your answer. – mbork Oct 16 '15 at 23:17

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 $g(x) = x\sin 1/x$.

• You know what? I got this confused with Swokowski Calculus. You are so right. – Todd Wilcox Oct 16 '15 at 19:57
• One should maybe add that it is also not discontinuous. It is neither continuous nor discontinuous, because talking about this doesn't make sense for points that are not in the domain. – philmcole Jan 16 '18 at 17:45