I've done some research, and I'm hoping someone can check me. My question was this:

Assume I have the function $f(x) = \frac{(x-3)(x+2)}{(x-3)}$, so it has removable discontinuity at $x = 3$. We remove that discontinuity with algebra: $f(x) = \frac{(x-3)(x+2)}{(x-3)} = (x+2)$. BUT, the graph of the first function has a hole at $x = 3$, and the graph of the second function is continuous everywhere. How can they be "equal" if one has a hole and the other does not?

I think that this is the answer:

Because the original function is undefined at the point $x = 3$, we have to restrict the domain to $\mathbb{R} \setminus 3$. And when we manipulate that function with algebra, the final result, $f(x) = (x + 2)$ is still using this restricted domain. So even though the function $f(x) = (x+2)$ would not have a hole if the domain were all of $\mathbb{R}$, we are sort of "imposing" a hole at $x = 3$ by continuing to throw that point out of the domain.

And then just to close the loop: Removing the removable discontinuity is useful because it allows us to "pretend" that we're working with a function that is everywhere continuous, which helps us easily find the limit. But the reality is that the function $f(x) = (x +2)$ is actually NOT continuous everywhere when we restrict the domain by throwing out the point 3. Or am I now taking things too far?

Thanks in advance!

EDIT: For anyone coming across this in the future, in addition to the excellent answers below, I also found this other question about the continuity of functions with removable discontinuities helpful.

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    $\begingroup$ You have two functions $g: \mathbb{R} \to \mathbb{R}$, and $f: \mathbb{R} \setminus \{3\} \to \mathbb{R}$, which are equal on all of $\mathbb{R} \setminus \{3\}$. The functions are not equal, since they have different domains. Actually both are continuous, but you have to pay attention to what "continuous" means on $\mathbb{R} \setminus \{3\}$. And yes, removing holes will not change how the function behaves, or how integrals involving the function work for example, and make things simpler $\endgroup$
    – Joppy
    Commented Mar 27, 2019 at 7:18
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    $\begingroup$ $x+2$ is continuous on $\mathbb R \setminus \{3\}$. $\endgroup$ Commented Mar 27, 2019 at 7:18
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    $\begingroup$ Yes you are right. In fact these two functions are equal on a common domain which is $\Bbb R-\{3\}$ but in $x=3$, $f(x)$ is not defined while $x+2$ is, so they can't be equal $\endgroup$ Commented Mar 27, 2019 at 7:28
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    $\begingroup$ Thank you all so much. Each of these comments added to my understanding. I learned all this as "cookbook" stuff years ago--just enough to follow the steps and get an A, not really caring to "understand" back then. I guess this has always been a "hole" in my knowledge. Pun intended :P $\endgroup$ Commented Mar 27, 2019 at 7:32
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    $\begingroup$ Continuity is a local phenomenon. A function is continuous or discontinous w.r.t a point, not the whole domain, though we can speak of continuity in the whole domain. $\endgroup$ Commented Mar 27, 2019 at 7:45

6 Answers 6


Two functions are typically defined to be equal if and only if they...

  • Share the same domain
  • Share the same codomain
  • Take on the same values for each input.

Thus, functions $f,g : S \to T$ for sets $S,T$ have $f=g$ if and only if $f(x) = g(x)$ for all $x$ in $S$.

For functions with holes, we typically restrict the domain by ensuring the values where the function is not defined at not included. For example, in the functions you have, you have

$$f(x) = \frac{(x-3)(x+2)}{(x-3)} \;\;\;\;\; g(x) = x+2$$

Are these equal? Yes, and no.

A function must be defined at all values of the domain. Thus, we can say $3$ is not in the domain of $f$ for sure. But we never specified otherwise the domains and codomains of these functions! Typically, unless stated otherwise, we often assume their domain to be $\Bbb R$ or $\Bbb C$, minus whatever points are causing problems - and of course, in such cases, $f \neq g$ since $f(3)$ is not defined, and thus $f$ normally has domain $\Bbb R \setminus \{3\}$ and $g$ generally has domain $\Bbb R$.

But that restriction is not necessary. For example, we could define the functions to be $f,g : \Bbb R \setminus \Bbb Q \to \Bbb R$. Notice that the domain of both functions are now all real numbers except rational numbers, i.e. the irrational numbers. This means $3$ is not in the domain of either function - and since that's the only "trouble spot," and the codomains are equal, and the values are equal at each point in the domain, $f=g$ here.

Or even more simply: we could have $\Bbb R \setminus \{3\}$ be the domain of $f$ and $g$ and again have equality! The key point in all this is that, just because $f$ or $g$ do attain defined values for certain inputs, doesn't mean they have to be in the domain.

In short, whether $f=g$ depends on your definitions of each. Under typical assumptions, $f \neq g$ in this case, but if we deviate from those assumptions even a little we don't necessarily have inequality.

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    $\begingroup$ I greatly appreciate this answer. The definition of function equality is very helpful. $\endgroup$ Commented Mar 27, 2019 at 7:47
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    $\begingroup$ Does the usual definition include sharing the codomain? I would happily define $f: \mathbb{R}\to\mathbb{R},\, f:x\mapsto x^2$ and $f: \mathbb{R}\to\mathbb{C},\, f:x\mapsto x^2$ to be the same function. $\endgroup$ Commented Mar 27, 2019 at 19:52
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    $\begingroup$ In my experience, yes, sharing the codomain is necessary. But much like altering the domain, you can easily manipulate the definition of the codomain (provided each $f(x)$ is in fact in the codomain). $\endgroup$ Commented Mar 27, 2019 at 22:29
  • $\begingroup$ @FedericoPoloni: The difference in the codomain is essential. This might become plausible when you consider $\mathbb{C}^{\mathbb{R}}=\{g:\mathbb{R}\to \mathbb{C}\}$, the set of all functions from $\mathbb{R}$ to $\mathbb{C}$. If you consider the functions as points in this space there are many more points near $f$. This implies you can for instance approximate $f$ by complex-valued functions which is not possible in $\mathbb{R}^{\mathbb{R}}$. $\endgroup$ Commented Mar 28, 2019 at 6:48
  • $\begingroup$ @MarkusScheuer I'm not too convinced by this example. Isn't this like arguing that the point 1 inside $[0,1]$ is different from the point 1 inside $\mathbb{R}$? $\endgroup$ Commented Mar 28, 2019 at 7:20

You are almost correct there!

The domain of the function matters, so for your example we have

$$f:\mathbb R\setminus\{3\}\rightarrow\mathbb R,~f(x)=\frac{(x-3)(x+2)}{x-3}.$$

You can think of it this way: we don't know yet if we have a removable discontinuity at $x=3$ and there might be a reason why we got this $(x-3)$ in the denominator, so we must exclude $3$ from our domain. Now our function $f$ is obviously continuous on its domain (it is a rational function and we know things about rational functions), and as we have excluded $3$ from our domain there is no point in asking if $f$ is continuous in $x=3$ (simply because it doesn't even exist there). Even when we simplify $$f(x)=\frac{(x-3)(x+2)}{x-3}=x+2$$ we still have the same domain because the domain does not change depending on our manipulations.

Now when it comes to asking wether we have a removable discontinuity we are actually asking the following: do we find a continuous function $g$ such that $$g:\mathbb R\rightarrow\mathbb R,~g(x)=\begin{cases} f(x),&x\neq 3 \\ c, &x=3 \end{cases}$$ So $g(x)=f(x)$ for all $x\in\mathbb R\setminus\{3\}$ (which is the domain of $f$) and for $x=3$ we are looking for a value to assign to $g(3)$ such that this "new function" $g$ is continuous. So because the domains of $f$ and $g$ are not equal the functions themselves are not equal, but for most purposes e.g. integration we can treat them as equal to make things easier. One example:

we want to calculate $\int\limits_{-5}^{2}\!f(x)\,\mathrm{d}x$. We then first have to discuss what we actually mean by that, as $f$ is not defined on $(-5,2)$ and after that we have an improper integral to solve, maybe split it up into two integrals...

Luckily one can show that in this case where we had a (single) removable discontinuity the following holds:


(This result can be extended e.g. it doesn't matter is we have a finite amount of removable discontinuities or $f(x)\neq g(x)$ for only finitely many $x$)

So working with $g$ makes this integration much easier which is why one often chooses to get rid of removable discontinuities and work with the new function $g$.

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    $\begingroup$ Really just a phenomenal answer. Thank you so much! $\endgroup$ Commented Mar 27, 2019 at 7:46
  • $\begingroup$ Perhaps you should clarify that your integral is not the standard Riemann integral, which cannot tolerate even a single undefined point. Other than that, good answer! $\endgroup$
    – user21820
    Commented Mar 27, 2019 at 12:30

As others have noted, the functions are equal on $\Bbb R\setminus\{3\}$, and $(x+2)$ is easier to work with in almost any respect. Yes, using $=$ in this case is an abuse of notation, but it's really common, and more or less universally accepted as a necessary evil.

However, there is a different perspective where $=$ is more correct, and that's if you see them not as functions, but as rational functions ("function" shouldn't be in this name, to be honest). In other words, as just fractions of abstract / formal polynomials, without worrying about any evaluation or function properties. Then they actually are equal, the same way $\frac62$ and $3$ are equal.

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    $\begingroup$ This is an interesting perspective. I'm going to have to do more research into rational functions. Thanks for the lead. $\endgroup$ Commented Mar 27, 2019 at 7:47
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    $\begingroup$ @1Teaches2Learn The field is called "Commutative algebra", and some keywords are "polynomial rings" and "ring of fractions". However, it is usually considered graduate level, so most material is going to be written accordingly. $\endgroup$
    – Arthur
    Commented Mar 27, 2019 at 7:51
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    $\begingroup$ @1Teaches2Learn: Rational functions are just what you get by treating polynomials as finite tuples encoding their coefficients, and considering 'fractions' of these polynomials, with two fractions $A/B$ and $C/D$ equivalent iff $A·D = B·C$. You get a field if you use polynomials over a field and forbid the denominator from being zero. Only if you wish to evaluate a rational function at some input, then you have to check that the denominator evaluated on the input is nonzero. $\endgroup$
    – user21820
    Commented Mar 27, 2019 at 12:40
  • $\begingroup$ $f_1=f_2$ is an abuse of notation, but $f_1(x)=f_2(x)$ is not. The former asserts that the functions are equal, while the latter asserts that the evaluations of the functions are equal. This gets into the fact that $f(x)$ is often referred to as a "function", when in fact $f$ is the function; $f(x)$ is the value of the function. $\endgroup$ Commented Mar 28, 2019 at 15:47
  • $\begingroup$ @Acccumulation But it's difficult to make that distinction when you don't have abstract function names, but rather the actual functional expressions to work with. That's what this question is mainly about as I see it: Not about the difference between $f_1(x) = f_2(x)$ versus $f_1 = f_2$, but about the exact same difference between $\frac{(x-3)(x+2)}{(x-3)} = x-2$ versus (for lack of better notation) $\frac{(x-3)(x+2)}{(x-3)} = x-2$. $\endgroup$
    – Arthur
    Commented Mar 28, 2019 at 15:51

They are equal as rational functions.

Both $$\frac{(x-3)(x+2)}{x-3}\quad\text{and}\quad x+2$$may be considered to be elements of the field $\mathbb Q(x)$ of "rational functons over $\mathbb Q$", and the two represent the same element of that field. So, when doing calculations in $\mathbb Q(x)$, it is, indeed, correct to write $$ \frac{(x-3)(x+2)}{x-3}=x+2 $$


One seldom writes the domain of a function, implicitly assigning the largest domain that is sensible in your context. For much algebra (especially graphing), trigonometry, and calculus, the largest domain of interest is the real numbers, $\mathbb{R}$, and the (implicit) maximal domains of functions are subsets of $\mathbb{R}$.

Two functions are equal if they have the same domain, range, and values on each point of the domain. (The range is a subset of the codomain. No one cares if you expand the codomain.) So $\frac{(x-3)(x+2)}{(x-3)}$ is not equal to $x + 2$ because they do not have the same domain: $\mathbb{R} \smallsetminus \{3\} \neq \mathbb{R}$.

Two functions are identical if they have the same values on each point of their common domain. The common domain of $\frac{(x-3)(x+2)}{(x-3)}$ and $x + 2$ is $(\mathbb{R} \smallsetminus \{3\}) \cap \mathbb{R} = \mathbb{R} \smallsetminus \{3\}$. Both functions agree on this common domain, so they are identical. "At any point you can evaluate both functions, they give the same answer, so there is no point of the common domain that can, through evaluation, show that the two functions are different." Two unequal identical functions are unequal "before you get to evaluations" -- that is, they are unequal in their domains or ranges.

A more extreme example: $\log x$ has domain $(0, \infty)$ and $\log( -x)$ has domain $(-\infty, 0)$. These two functions are not equal; they do not have the same domain. Because the intersection of their domains is empty, their common domain is empty and they are (vacuously) identical. (Here, "vacuously" means "there is literally nothing to check because the common domain has nothing in it, like a vacuum".)

When proving identities, we only use this weaker notion of equivalence of functions.

Regarding continuity: "$f$ is continuous" means that $f$ is continuous on each point of its domain. If you delete a point from its domain, you have relaxed the conditions imposed by continuity. "$f$ is continuous at the point $x$" means $$ \lim_{t \rightarrow x} f(t) = f(x) $$ (where the indicated limit must exist). In considering the limit, $t$ is restricted to only take values in the domain of the function.

(Most people don't talk about continuity of functions at isolated points of their domains. With the above definition, a function is continuous at any isolated point of its domain because the limit becomes vacuous: there are no points in the domain near the isolated point except for the isolated point, so there is nothing to check. It is not the case that the limit fails to exist. In a rigorous setting, you would define continuity as "for all $\varepsilon > 0$, there exists a $\delta$ such that for all $t$ such that $|x-t| < \delta$, we have $|f(x) - f(t)| < \varepsilon$". For an isolated point, once $\delta$ is small enough, the only choice for $t$ is $t = x$, and we have $|f(x) - f(t)| = 0$, which is definitely less than $\varepsilon$.)

Continuous functions are nice. For instance, you likely have a theorem that says continuous functions are (Riemann) integrable (on non-infinite intervals of integration). If you have a function that has a removable discontinuity, then it is identical (but not equal) to a continuous function whose domain includes the ordinate (first coordinate) of the removable discontinuity. Any value you can get from the first function, you can also get from the second function. This means any particular Riemann sum (that is, any particular choice of partition and sample points) that can be evaluated using the discontinuous function has the same value as the same sum using the identical, continuous function. Riemann sums that sampled the removable discontinuity did not exist, so prevented the existence of the limit as the diameter of the partition went to zero. The identical function sidesteps this problem by supplying the limit of the function as it approaches the removable discontinuity, so the Riemann sum using the "filled in" function exists and you can integrate it. (Normally, one talks about an improper integral, splitting the interval of integration into pieces that avoid little intervals around the points of discontinuity, then taking limits as the little intervals shrink to zero. If the discontinuity is removable, one can show that these two methods of sneaking up on the removable discontinuities give the same result.)

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    $\begingroup$ Actually the codomain is part of the definition of the function, so it does matter. A surjective function fails to be surjective if you expand the codomain. $\endgroup$ Commented Mar 28, 2019 at 0:09
  • $\begingroup$ @MattSamuel : You have explained exactly why codomain is useless and is the wrong data for the definition of the function. All functions are surjective on their ranges. $\endgroup$ Commented Mar 28, 2019 at 0:10
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    $\begingroup$ But it sometimes matters whether a certain naturally defined function is surjective on its codomain. The codomain is not necessarily arbitrary. If you're taking about real valued functions of a real variable, sure, it doesn't matter very much. But even as soon as you get to complex analysis it matters. Certainly in higher math it matters a lot. $\endgroup$ Commented Mar 28, 2019 at 0:14
  • $\begingroup$ @MattSamuel : If I hand you a function modelled in the usual way (a set of of domain-image pairs), you cannot possibly know the codomain, unless you mark every element with its defining set, so that $1_\mathbb{Z} \neq 1_\mathbb{R}$. All the model can provide is the range (union of images). Asserting a codomain is creating a fiction for which the function is an inadequate witness unless we force our elements to not satisfy natural inclusions. $\endgroup$ Commented Mar 28, 2019 at 0:19

You're assuming that f(x) is not defined at x=3 because f(3) produces 0/0, and because this expression is ordinarily undefined then you're assuming that f(3) is undefined. But if you can rationally give a definition to this expression then we're on solid ground. A rational value can be given to this expression in this case by using l'Hôpital's rule, and so f(3) is defined.

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    $\begingroup$ There are several correct answers to this subtle question. Yours is not. Please read them. What you have observed is that it's easy to change the function slightly to make it continuous (by increasing the domain). And L'Hopital has nothing to do with the construction. $\endgroup$ Commented Apr 4, 2019 at 2:50
  • $\begingroup$ Your answer is wrong, simply because a function definition consists of $3$ things, the relevant one here being the domain. The division by zero means that the point cannot be in the domain of the function. That the limit exists at that point is irrelevant. $\endgroup$ Commented May 11, 2019 at 10:02
  • $\begingroup$ That's not to say you can't define $f(3)=5$ as a separate case in the definition of the underlying mapping and use the $(x-3)(x+2)/(x-3)$. You can absolutely can do that, and even retain continuity of $f$. But then you change the definition of $f$. As presented, $f$ has the definition that prevents $3$ from being in the domain. You could make a function, say, $\overline f$ such that $\overline f(x)=f(x) \; \forall x \ne 3$ and $\overline f(3) = 5$. This is however not equal to $f$ since $f$'s definition vehemently prevents definition at $x=3$, however nice said definition could be. $\endgroup$ Commented May 11, 2019 at 10:08
  • $\begingroup$ The reason that $\overline f \ne f$ being that $dom(f) \ne dom(\overline f)$ under the usual assumptions. We can define these domains mostly as we please, up to $f$ not having $3$ in it. We could define $\overline f, f$ on the irrationals and have $\overline f = f$ since the domains are equal ($3$ is not irrational). But if we want the domain of all complex numbers, we'd have to delete $3$ from the domain of $f$ to prevent the issue. $f=g$ iff $dom(f)=dom(g), cod(f)=cod(g),$ and $f(x)=g(x) \; \forall x\in dom(f) = dom(g)$. $\endgroup$ Commented May 11, 2019 at 10:11
  • $\begingroup$ Therein lies the subtlety Ethan was referring to. Typical function equality has a very precise definition. Whether $f, \overline f$ are equal depends on the assumptions we make in this problem, namely the definition of the domains. Sometimes they're equal, sometimes they're not. What is definitely true however is that limiting behavior does not determine the domain of a function. You can find any real number limit through a sequence of rationals, but that doesn't mean that the domains of functions defined exclusively on the rationals have domain $\Bbb R$, just sometimes can be nicely extended. $\endgroup$ Commented May 11, 2019 at 10:16

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