# is the reciprocal function continuous

I'm curious the views of those who teach calculus.

As you know the continuity of a function at a point is defined in terms of the limit in the typical course. I'd like to ask a pair of questions:

1. Consider $f(x)=1/x$. Is $f$ continuous ?
2. Let $\text{dom}(g) = (-\infty, 0 ) \cup (0, \infty)$ where $g(x)=1/x$ for each $x \in \text{dom}(g)$. Is $g$ continuous?

I don't want to say much more initially as I'd rather not influence the answers at the outset. The audience I have in mind is a mixed math majors / science majors calculus course at the university and time is not a large issue.

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Whether or not this is a reasonable pair of questions depends on whether students have been taught the definition of a continuous function. Also if the definition is that a function is continuous if it is continuous at every point in its domain, then the students will need to be able to figure out the domain of $f$ from $f$'s definition. If these are things you've taught the students, then the question pair is reasonable. If what you've taught them differs from what they might have been taught in precalculus, then the difference is something that should probably be emphasized in lectures. – Theodore Norvell Aug 9 '14 at 21:25

I think the point of the question is that although there is no mathematical ambiguity, in different circumstances one would give different answers. So:

I. From the perspective of a freshman calculus textbook: well, I know my enemy. The textbook answer to (1) is "A function is 'continuous' if it is continuous at $c$ for every real number $c$. A function is continuous at $c$ if it is defined at $c$, if $\lim_{x \rightarrow c} f(x) = L$ exists and $L = f(c)$. Since $f(x)$ is not defined at $0$, it is not continuous there." Yikes. To be fair, most calculus textbooks would classify the discontinuity at $0$ as not being removable.

I feel ever so slightly worried that I've unfairly caricatured this addle-pated view of continuity. I know that freshman calculus textbooks would give the answer "No, because it is not defined at $0$" to (1). As written, it sounds like they would give a similar answer to the continuity of, say, $f(x) = \sqrt{x-1}$, and I think that they may not be as legalistic as I implied above that a function can only be called continuous if it is defined at all real numbers: sometimes being defined on a closed interval is regarded as sufficient. I think question (2) outfoxes the freshman calculus text: even by asking it you show understanding that the continuity of a function depends upon what domain you take. The calculus texts are not clear on this.

II. From the perspective of a mathematician: (1) I would ask, "Well, what is the domain and codomain of the function? If the function is meant to take real or complex values, then it is not defined at $0$. As a function from $\mathbb{R} \setminus \{0\}$ to $\mathbb{R}$ or from $\mathbb{C} \setminus \{0\}$ to $\mathbb{C}$, it is continuous (and more...). The function cannot be extended continuously to $0$, e.g. because it is unbounded in every neighborhood of $0$. However, it extends continuously [and more...] to a function on the Riemann sphere: $f(0) = \infty$." (2) Yes, of course $g$ is continuous [and more...].

III. If a freshman calculus student asked me this in class:

(1) "There are some tricky issues in continuity for functions which are not defined at isolated points. Our intuitive notion of continuity is that the graph is a nice unbroken curve. [Draws the graph of f] As you can see, in this case the graph is not nice at $0$. Now in fact the function is not even defined at $0$. Whether a function is continuous at a point where it is not defined is a bit legalistic! More pertinently, sometimes functions which are not defined at an isolated point can be defined at the point so as to be continuous there. This happens exactly when the limit as you approach the point exists and is called a 'removable discontinuity'. However, in this case, $\lim_{x \rightarrow 0^{-}} f(x) = -\infty$ and $\lim_{x \rightarrow 0^{+}} f(x) = \infty$ so it's pretty bad: each one-sided limit is infinite, but the signs are different, so we can't even say that the overall limit is $\pm \infty$ [even if we could, since $\pm \infty$ are not real numbers, the limit would not exist]. The upshot is that there is no "good" way to define $f(x) = \frac{1}{x}$ at $0$."

(2) "Yes, the function is definitely continuous on the given domain, as you can see from the graph. Beware though: the given domain is not an interval but rather a union of two non-overlapping intervals. Be very careful with this: most of the big theorems in calculus concern continuous functions defined on an interval. When the domain is 'more than one interval', some funny things can happen. [Depending on where we are in freshman calculus...] For instance the intermediate value theorem fails here: the function takes negative values and positive values but never takes the value $0$. Also one antiderivative of the function is $\log |x|$, but not every antiderivative is of the form $\log |x| + C$! [Maybe say more about this if someone seems interested; it is a tricky point.] It may be more useful to think of the function $g$ as really being 'two different functions', one defined for negative values and one defined for positive values."

Added: The most general antiderivative of $g$ seems like a tricky point indeed, since more than one person has asked about it in the comments. It may be worth explaining the answer. The uniqueness of antiderivatives up to a constant is equivalent to the Zero Velocity Theorem: if $F$ is differentiable and $F' = 0$ (identically) then $F = C$ is constant. This in turn is a consequence of the Mean Value Theorem, and the hypothesis that the domain be an interval is critical: more generally any locally constant function has derivative zero, and if the domain has $N$ connected components, the vector space of locally constant functions has dimension $N$. In this case the domain of $g$ consists of two disjoint intervals, so the most general antiderivative is $G(x) =$ $\log(x) + C_1$, $x > 0$
$\log(-x) + C_2$, $x <0$
for any two $C_1,C_2 \in \mathbb{R}$.

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one antiderivative of the function is $\log |x|$, but not every antiderivative is of the form $\log |x| + C$. Is there any other antiderivatives? – whatever Aug 9 '14 at 10:25
@PeteL.Clark OK, that's obvious. I thought there's an antiderivative that doesn't use $\ln x$ . – whatever Aug 9 '14 at 14:58
+1 especially for how a mathematician would salvage the first question. – djechlin Aug 9 '14 at 22:27
@metacompactness By the fundamental theorem of calculus if $F(x)$ is the antiderivative of $f(x)$ then any other antiderivative can only differ by a constant. So no, there's no other antiderivatives. – Kristoffer Ryhl Aug 10 '14 at 9:02
@Darksonn: The Fundamental Theorem of Calculus applies separately on each interval. Thus the most general antiderivative of $g$ is $F(x) = \log x + C_1$ for $x > 0$ and $\log (-x) + C_2$ for $x < 0$, where $C_1,C_2$ are two arbitrary constants. (metacompactness and I had some previous exchanges about this which we both deleted.) – Pete L. Clark Aug 10 '14 at 16:20

In this situation, I'd claim there's the "prior error" of seemingly requiring a boolean answer, yes-or-no. Surely we're not so much interested in pranking the students by giving them a function whose domain is not topologically connected, which is continuous on each connected component, but not "continuous" in an obvious, intuitive sense "across the gap". The situation is clear, because everyone in Calc I knows the graph of this function. What is not clear is the semantics of a formal definition, whose pitfalls will not have caught the students' attention... and maybe don't merit it, besides.

That is, "$f(x)=1/x$ has a problem at $x=0$, but otherwise is fine".

("What about $1/x^2$?")

That is, I'd propose not over-burdening a boolean "continuity-or-not" with this situation in Calc I. It's not so much a mathematical issue as a semantic one, which oughtn't be a primary point, in my opinion.

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I tend to agree with your viewpoint about not trapping students. So, if this was a major point in the course and it was my goal to "get" them on it I could see it. But, the reason I discuss this issue (when I do) is to bring attention to the limitations of the definitions. In the same course, at other points, I am quite careful about the $\epsilon \delta$ proofs, so, it seems lopsided to just abandon logical clarity for the sake of intuition at this point. I mean, intuitively who needs the $\epsilon \delta$ proofs? Of course, the issue can be avoided by focusing on intervals. – James S. Cook Aug 12 '14 at 4:08

I think most(?) textbooks (Stewart, at least) do not define the statement, "$f$ is continuous." They define $f$ being continuous at a number and $f$ being continuous over an interval. So questions 1 and 2 would not be well-posed with respect to such a textbook. One has to pick a number or an interval first and then ask about the continuity of $f$ at the number or over the interval.

A harder question to answer is whether $h(x) = \sqrt{x^4-x^2}$ is continuous at $x = 0$.

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I don't really know, but possibly citing a calculus textbook writer (with no other mathematical distinctions whatsoever) as an authority on such a matter is perceived as not to the point. All the worse if this "authority" declares a common and reasonable question "ill-posed", I'd guess. Even if the original question is somewhat "asking for trouble", simply declaring it illegitimate by authority of Stewart does not really respond, eh? In real life, "$f$ is continuous" is a commonly-heard, commonly-understood phrase, after all. – paul garrett Aug 11 '14 at 22:00
@paulgarrett Others doing the guesswork for why there's a down-vote is potentially helpful, but it would be much better if dvers added an explanation! Clandestine disagreement serves only to (as ME2 says) discourage. – Benjamin Dickman Aug 11 '14 at 22:21
@BenjaminDickman, yes, I agree, but/and we can imagine many situation in which downvoters might be inhibited... – paul garrett Aug 11 '14 at 23:00
Michael, you may have suffered from the tl;dr syndrome, that Rory D's answer was not read, hence, not reacted-to, because it was "too long", while yours was more pithy. :) In any case, I myself give no weight whatsoever to calculus textbook authors who are no more than that... Even if we concede that much "teaching of calculus" is merely filtering for other undergrad majors (in the U.S.), this does not argue to me that we should make things up, make artificial rules, make non-physical conventions, or deny obvious intuition. True, if we admit that nearly all humans can "do math"... – paul garrett Aug 12 '14 at 0:25
... then we confound the filtering mechanism, and sorta put ourselves outta jobs... not to mention pretending to delegitimize the highly-formal or highly-stylized mathematics that is used as further filter/hurdle. Nevertheless, I do claim that we can "filter" on more important traits than adherence to semi-random, highly artificial, semantic distinctions that do not correspond to worthwhile distinctions in real life. – paul garrett Aug 12 '14 at 0:27

This is my experience as a calculus TA for the last 4 years, and I have had no active role in setting the curriculums I'm talking about. The "continuity of a function" may be defined and explained thoroughly in an advanced honors rigorous calculus class, but in the trenches that are the first year calculus at most first year universities it does not play a large role. There are essentially 3 options for defining continuity for these students.

1. A function is continuous if its graph can be drawn without picking up the pen. This is the most popular method (in what I've observed) , and it leads to the fallacy you are hinting at. Still, this definition (whether flawed or not) is understandable to calculus students and is of the kind of thinking that mathematicians use.

2. $\lim_{x\rightarrow a} f(x)=f(a)$. This definition is in every calculus text and is of course entirely correct. The issue is that this definiton has the giant black box of a "limit", which many students simply will not understand unless they are forced to (through HW or lectures). At least in my university the pressure to skip over these more difficult and harder to test aspects of calculus in favor of a larger more impressive-looking curriculum guarantees that they will not ever get experience in things like limits.

3. $\forall\epsilon\ \exists\delta$ s.t $|f(x)-f(x_0)|<\epsilon\ \forall|x-x_0|<\delta$. Again if an effort is made to teach this kind of a definition, it can be taught to first years. There simply isn't time for that in the curriculums I have seen.

The goal of most calculus curriculums I have been party to is to get the students calculating integrals and derivatives of elementary functions. What is done after that is not necessairly constant, but almost all of these classes are in a sprint to get to this material. Generally what I think the very best students in such a class understand that continuity of a function allows us to apply the fundamental theorem of calculus to that function and perhaps the first definition I've given. There is a huge expanse of material to cover in a couple of semesters and foundations generally get short shrift. Saying that, without some power to change the curriculum, I cannot expect students to try and understand difficult concepts when they will be given tests that completely ignore these concepts.

EDIT: If this seems as an indirect answer it somewhat is. I understand what continuity means if it were said in a research seminar, but I choose to reinforce definition 1 to calculus students then the others as I think it is much more likely to be understood.

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The definitions you give are silent on endpoints. Is $\sqrt{x}$ continuous at $x=0$? But, I understand, it's hard to cover all the cases when we're mostly teaching algebra which has yet to be mastered. A compromise I've found is to write the two-sided limit definition and mutter in class that we replace with appropriated one-sided limits for points in the domain. The graph-drawing technique is hard to violate if you add the critera "on a connected domain" aka interval. – James S. Cook Aug 9 '14 at 14:17

I thought it might be interesting to look at a variety of different Calculus textbooks to see how they handle the terminology. Here is an unscientific survey of the half-dozen different Calc textbooks I have on my bookshelf right now:

Stewart, Single Variable Calculus, 4e (both Early Transcendentals and Late Transcendentals versions) and 5e:

Stewart defines the phrase continuous at a number a in the usual way; the text also defines continuous from the right at a and continuous from the left at a. Armed with these definitions, the text defines continuous on an interval, with one-sided limits used at the endpoints of the interval if appropriate in context.

With respect to the specific question of the OP, Stewart's Theorem 5 states that "Any rational function is continuous wherever it is defined; that is, it is continuous on its domain."

The phrase "a continuous function" (without qualifiers like "at a point" or "on an interval") is used in the exposition, but never in the definitions, theorems, proofs, or exercises. The phrase is used solely as a kind of informal shorthand that makes it possible to refer to several different types of continuity at once, for example in statements like "a sum or product of continuous functions is continuous". A question like "Is the reciprocal function continuous?" would not be found in this text; however, "Is the reciprocal function continuous on its domain?" and "At what points is the reciprocal function discontinuous?" would be legitimate.

Gillett, Introduction to Calculus and Analytic Geometry, 4e

Like Stewart, Gillett defines "continuous at $x_0$". This text also defines "continuous in S" (where S is a subset of the domain). (Interestingly the definition does not impose any restrictions on what types of subsets may be considered, but every example seems to be a countable union of intervals.) The phrase "continuous in its domain" comes next, and Gillett has as Example 6 that "The reciprocal function is continuous in its domain."

Regarding the terminology "a continuous function", Gillett has the following remark immediately following Example 6:

When a function is continuous in its domain, many writers simply call it continuous, using the global term (unqualified)… This is not unreasonable… but it can be misleading. To say that f(x) = 1/x is continuous (without any qualification) is to risk overlooking the infinite jump at x = 0. It also sounds paradoxical to say "f(x) is continuous" in one breath and "f(x) is discontinuous at x=0" in the next. That is why we prefer the language "continuous in its domain." It leaves the door open for further remarks.

So Gillett explicitly rejects the phrase "a continuous function" without specifying a point or a subset of its domain, and justifies this using the reciprocal function as its example of why that usage would be poor.

Moise, Calculus: Part I.

This text (from 1966), like the more modern examples, starts by defining "continuous at $x_0$". Unlike Stewart and Gillett, Moise continues by defining "If $f$ is continuous at every point $x_0$ of its domain, then we say that $f$ is continuous." So it uses precisely the terminology that Gillett critiques.

As an example of how continuous functions can be combined to produce more continuous functions, Moise considers the example of $f(x) = x^2 + 1$ and $h(x) = x^2 - 1$. The text states that $f+h$ and $fh$ are both "continuous" (note the absence of any qualifier) and then says that $f/h$ is continuous except at $1$ and $-1$. The example continues, "Of course, at $x=1$ and $x = -1$ it is not just continuity that breaks down: the quotient function is not even defined at these points, because the denominator of $f/h$ becomes $0$." Oddly enough, Moise stops short of saying that $f/h$ is a "continuous function", despite the fact that it would be completely in keeping with his own usage conventions to do so. Presumably that is because, as Gillett noted in his remark quoted above, it would seem weird to say that "$f/h$ is continuous" and then also say that "$f/h$ is discontinuous at $1$ and $-1$".

Zill (1993), Calculus (3e)

Like all the others, Zill begins by defining "continuity at a point". Zill states that that rational functions are discontinuous at points where they are not defined. The text defines continuity on an interval (with "continuous from the right/left" used at end points as appropriate).

With respect to the usage "continuous function", Zill says:

Functions that are continuous on $(-\infty, \infty)$ are said to be "continuous everywhere" or simply "continuous".

So Zill reserves the phrase "continuous function" for use only with functions whose domain is all of $\mathbb{R}$. On the other hand (and somewhat in contradiction to this explicit definition), "continuous function" is used throughout the exposition (but never in the theorems!) as a shorthand, much in the same way that Stewart does.

So, to summarize, I'm going to go out on a limb here and, based on the single example of the 1966 textbook, suggest that the use of the phrase "continuous function" to mean "function that is continuous at each point of its domain" might be an artifact of an earlier generation of textbook authors. More recent texts either avoid giving the phrase any technical meaning at all, using it instead as an informal shorthand, or (in the case of Zill) define it to mean "continuous on all of $\mathbb{R}$".

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I cover three different concepts of continuity in my high-school calculus class (separate from advanced concepts such as uniform continuity). This vocabulary is based on the AP textbook "Calculus: Graphical, Numerical, Algebraic" by Ross L. Finney et al., pages 75 and 77.

1. A function $y=f(x)$ is continuous at an interior point $c$ of its domain if $$\mathop {\lim }\limits_{x \to c} f(x) = f(c)$$ And similarly for an endpoint of an interval, using a one-sided limit. A function is not continuous at any point not in its domain. Hence your reciprocal function is continuous at every value of $x$ other than $x=0$, where it is discontinuous.

2. A function is continuous on an interval if and only if it is continuous at every point of the interval. Your reciprocal function is continuous on every interval not containing $x=0$.

3. A continuous function is one that is continuous at every point in its domain. Hence, the answer to your first question is: for $f(x)=\frac1x$, $f$ is a continuous function. And so is your function $g$.

These definitions are not perfect but they are indeed helpful in my teaching. I can use theorems about continuous functions without being concerned about points not in the domain.

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See, this to be is logical, but it makes some of my colleagues uneasy as it disagrees with precalculus discussions where the pen-drawing definition makes $y=1/x$ a discontinuous graph. I was discussing this with brother and we agreed to disagree. For him, the nineteenth century conflation of the function with its formula and the intuitive appeal is not worth abandoning for the sake of matching the technical defn. of continuity we give in later courses. I'm generally a fan of doing things right the first time, so I like your approach. But, I see @Pete L. Clark shares my brothers view. – James S. Cook Aug 9 '14 at 14:21
To be clear, I don't think it is wrong to teach $1/x$ is discontinuous in a class where domains are not emphasized. However, if serious discussion is made of domains of functions then it seems out of place to me to say $1/x$ is discontinuous at $x=0$. Unless, you give a definition of discontinuity which explicitly defines discontinuity for points outside the domain. – James S. Cook Aug 9 '14 at 14:23
@JamesS.Cook: In my class I emphasize that the graph of a continuous function can be drawn with a pen kept on the paper through any point on the graph but not necessarily for the entire graph. This has worked best for me as a compromise in the various, somewhat conflicting ideas about continuity. I do see the point of what you say. – Rory Daulton Aug 9 '14 at 15:13