# Why are removable discontinuities even discontinuities at all?

If I have, for example, the function $$f(x)=\frac{x^2+x-6}{x-2}$$ there will be a removable discontinuity at $x=2$, yes?

Why does this discontinuity exist at all if the function can be simplified to $f(x)=x+3$?

I suppose the answer is that you can't simplify it because you can't divide by something that could potentially equal $0$.

But what if you start with $f(x)=x+3$? What's stopping you from multiplying it by $1$ in the form $\frac{x-2}{x-2}$? Can multiplying by $1$ really introduce a discontinuity to the function?

• Since it is a removable discontinuity which returns a reduced form of the original function, it still remains that the domain of the original function is still in effect. – user60887 Nov 12 '15 at 0:59
• A function $f$ is continuous at a point $c$ because it satisfies three things: 1) $f(c)$ exists, 2) $\lim\limits_{x\to c}f(x)$ exists, and 3) $f(c)=\lim\limits_{x\to c}f(x)=f(c)$. A function is not continuous at a point $c$ if any one of these three conditions fail. If 1) or 3) but not 2) fails at a point $c$, the continuity is removable. If 2) fails at a point $c$, the discontinuity is non-removable. I think this is simply the definition of removable/non-removable. Also, note $f(x)=x+3$ and $f(x)=\frac{x^2+2-6}{x-2}$ are different functions. Emphasis on the word ''functions''. – user31415926535 Nov 12 '15 at 1:09
• You'd be multiplying by an expression that’s undefined at $2$, so you're introducing a discontinuity that wasn't there in $x+3$. This sort of thing is the error that lurks in many popular “false proofs." – amd Nov 12 '15 at 1:11
• @user31415926535 By that logic, isn't any function discontinuous because it has a discontinuity at every value not in its domain? So $h : \mathbb{R} \rightarrow \mathbb{R} : x \mapsto x$ is discontinuous at "potato". I believe discontinuity only makes sense at values in the domain so 2 is not a discontinuity of $f$ and therefore $f$ is continuous. – Reinstate Monica Nov 12 '15 at 19:25
• Removable singularity is a better term to use, since a function which is not defined at $x=2$ isn't discontinuous at $x=2$ to begin with. See the last paragraph in this section of the Wikipedia article: en.wikipedia.org/wiki/… – Hans Lundmark Aug 14 '17 at 10:12

It comes down to what we mean by a function. In specifying the function, we need to specify not just the "rule", but also the domain, i.e., what are you allowed to put into the function?

In particular, to answer your question directly: if we want $f(x)=x+3$ to be defined at $x=2$, then we aren't allowed to say things like $$f(x)=f(x)\cdot\dfrac{x-2}{x-2},\,''$$ because the expression on the right is not defined at $x=2$.

I think it might be the norm to brush over this kind of thing in introductory calculus courses, because in such a course we tend to assume that the function is defined "everywhere it makes sense" (a frame of mind which is convenient, but carries many problems with it besides just this).

To be a bit more explanatory, if you just say the function $f$ is specified by $$f(x) = \frac{x^{2}+x-6}{x-2},$$ then you haven't specified the function enough to be satisfactory for a really rigorous treatment. In particular, is your function defined at $x=2$, or is it not? Given the way that you have written it down, I would assume that it is not, since, as you point out, we cannot divide by $0$.

The way we deal with this in a rigorous manner is to require the definition of the function to specify the domain. For example, you could say that $f$ has domain $\mathbb{R}$ and is given by $$f(x) = \begin{cases}\dfrac{x^{2}+x-6}{x-2}, & x\neq2, \\ 5, & x=2.\end{cases}$$ Note that the only reason this makes sense is that your function is specified for all $x$ in the domain (in particular, it is clear that there is, for this value of $f(2)$, no discontinuity at all). If we had said "$f$ has domain $\mathbb{R}$ and is given by $$f(x) = \dfrac{x^{2}+x-6}{x-2}$$ for all $x$" then we would run into problems: we claim that $f$ has domain $\mathbb{R}$, but $2\in\mathbb{R}$ and $f(2)$ is not defined.

By contrast, you could specify your function as the function with, for example, domain $[0,1]$ given by $$f(x)=\dfrac{x^{2}+x-6}{x-2},$$ when we see that there is, in a sense, no discontinuity at $x=2$ since $f(2)$ has not been defined. In this case, it makes sense to say $$f(x)=f(x)\cdot\dfrac{x-2}{x-2},$$ since this expression now makes sense for all of the values in the domain of the function. It also now makes sense to make the cancellation and say $f(x)=x+3$, because our knowledge of the domain of the function ensures that we are not dividing by $0$.

What's stopping you from multiplying [a continuous function] by $1$ in the form $\frac{x−2}{x−2}$? Can multiplying by $1$ really introduce a discontinuity to the function?

Ah, but $\frac{x−2}{x−2}$ is not equal to $1$. Or, to be more, precise, it is only equal to $1$ for values of $x$ that are different from $2$. If $x=2$, then $\frac{x−2}{x−2}$ is undefined.

And that's more or less the answer to your question: If you take a given function $f(x)$ that is continuous on all of $\mathbb{R}$, and you multiply it by a function $g(x)$ that is equal to $1$ everywhere except at a single isolated point, at which point $g(x)$ is undefined, then the product $f(x)g(x)$ will be exactly the same as $f(x)$ everywhere except at that point, where it, too, will be undefined.

Neither the function $$f(x)=\frac{x^2+x-6}{x-2}$$ nor the function $g(x)=x+3$ has a discontinuity at $x=2.$ In the first case, it is because $f$ is not defined at $x=2,$ so is not (continuous or discontinuous) there. In the latter case, it is continuous there.

Now, consider the family of functions $h_\alpha(x)$ given by $$h_\alpha(x)=\begin{cases}x+3 & x\ne 2\\\alpha & x=2,\end{cases}$$ where $\alpha$ is a real constant. The function $h_5(x)$ is simply $g(x),$ expressed in a more complicated form. For any real $\alpha\ne5,$ $h_\alpha(x)$ has a removable discontinuity at $x=2.$ What this means is that we can make $h_\alpha(x)$ into a (different) function that is continuous at $x=2$ by choosing a different value for $h_\alpha(2).$

Consider the function $i(x)=\frac{x-2}{x-2},$ now. Note that $i(2)$ is undefined, and so $i(x)$ doesn't make sense unless $x\ne2.$ And, of course, for $x\ne2,$ we have $i(x)=1.$ Now, in order to multiply $g(x)$ by $i(x),$ both of these have to be defined, and so we have to require that $x\ne 2$ once again. So, while $g(x)$ is defined everywhere, $f(x)=g(x)\cdot i(x)$ is not.

Added: It's worth noting that the sources to which you linked in the comments are not using the standard definition. (However, this abuse of terminology is far from uncommon, even among textbooks!) Moreover, their definitions don't even agree with each other!

The first source (or, at least, what I can read of it) seems to adhere to an informal definition that suggests the following formal definition:

A function $f$ has a removable discontinuity at a point $x_0$ if there is a number $L$ such that $\lim_{x\searrow x_0}f(x)=L=\lim_{x\nearrow x_0}$ and either (i) $f(x_0)$ is undefined or (ii) $f(x_0)$ is defined and is not equal to $L.$ In the former case, we say that $f$ has a hole at $x_0.$ In the latter case, we say that $f$ has a created discontinuity at $x_0$.

The terminology "hole" (and this definition of it) is fairly standard. However, the definition of "created discontinuity" (a term I've never seen before) is exactly the standard definition of "removable discontinuity"!

The sketched graph provided there gives an example of a hole, and the source also seems to refer to a graph giving an example of a "created discontinuity," but it isn't visible to me.

The second source, on the other hand, conflates the terms "hole" and "removable discontinuity," as opposed to your first source (which treats holes as special cases of removable discontinuities) and the standard usages (in which holes and removable discontinuities are mutually exclusive). Moreover, its attempt at a formal definition is at best informal, and at worst self-contradictory!

Formally, a removable discontinuity is one at which the limit of the function exists but does not equal the value of the function at that point; this may be because the function does not exist at that point.

The part preceding the semicolon is the usual definition, as, by speaking of "the value of the function at that point," it is implied that the function takes a unique value at that point! However, this is directly contradicted by the part after the semicolon. Thus, if we attempt to make it formal, we fail. If we leave it informal, we can ignore the implication and thereby avoid the contradiction (but not potential confusion).

The graph provided there gives an example of a removable discontinuity (as usually defined), but not of a hole (as usually defined).

• That wasn't my understanding of what a removable discontinuity is. – Kyle Delaney Nov 13 '15 at 19:44
• How was "removable discontinuity" defined for you? – Cameron Buie Nov 13 '15 at 22:04
• The way my question and Will R's answer implies. It's a point where the function doesn't fit or is undefined, often because the function evaluates to $0/0$ at that point. study.com/academy/lesson/… mathwords.com/r/removable_discontinuity.htm – Kyle Delaney Nov 18 '15 at 20:37
• You seem to be misreading Will R's answer. At no point does he use the words "removable discontinuity," nor suggest that $$f(x) = \dfrac{x^{2}+x-6}{x-2}$$ has a removable discontinuity. In fact, in every case he discusses, he says that $f$ does not have a discontinuity at $x=2.$ I have edited my question to address the definitions given by the sources you've linked. – Cameron Buie Nov 19 '15 at 0:15
• So instead of saying removable discontinuity I should just say hole? Is hole understood to mean a single point where the function is undefined due to $0$ being divided by $0$? Could hole not also refer to a larger gap in the function, say if the function isn't defined between $-1$ and $1$? – Kyle Delaney Nov 20 '15 at 1:34

Simply crossing out common factors from the original function does not change the fact the x cannot exist at x=2...

By crossing out the common factor of (x-2) from both the numerator and the denominator, you need to state that x cannot equal 2 (otherwise division of the common factor cannot take place; cannot divide by zero). This accounts for the removable discontinuity at x=2

Also, when you multiply by $(x-2)/(x-2)$, you are multiplying by 1, except for when x equals 2... then you are really multiplying by something that is undefined.