# Teaching Asymptotes

Yes, I've read a number of definitions

In analytic geometry, an asymptote of a curve is a line such that the distance between the curve and the line approaches zero as they tend to infinity. Some sources include the requirement that the curve may not cross the line infinitely often, but this is unusual for modern authors.

And for most examples, such as $y=1/x$, it seems clear what's going on, and how to discuss the asymptote. Yesterday, a student asked me about this function - Only after I told her it was a discontinuous function, and when drawing it, be certain to show that at x=0, both lines should be an open circle indicating "not a point", did she tell me the teacher said that $y=1$ and $y=-1$ were asymptotes. I think there's a mistake here as there's nothing approaching the lines, the lines are the equation. Just as a first order binomial, $y=x+2$, is just a line, no asymptotes there.

This prompted the follow on - And the question of whether today's usage allows crossing to occur. In this example, the distance itself, from first definition, will cross zero, infinitely many times. Looking at the quoted definition, what does "modern authors" mean? Do I need to be concerned that an older book will call the X axis of second graph "Not Asymptote", but a newer one is fine?

• In my opinion as a mathematician, "asymptote" is a somewhat informal term, and like most informal terms they break when stretched too much or when you look too hard at them. I tend only to use it in settings where it's traditionally used, such as describing hyperbolas, or graphs of nonlinear rational functions. In other settings, I would use other terms. I often use other terms even in these settings. – user797 Sep 17 '16 at 12:56

Asymptote - if $f(x)-g(x)\to0$ as $x\to a$, then $f(x)$ is asymptotic to $g(x)$ as $x\to a$. Similarly, $g(x)$ is asymptotic to $f(x)$ as $x\to a$. (IMO)

Personally, I do not see the problem with intercepting any amount of times. If you wish to modify the definition of an asymptote to fit this, I would probably use limit superior and limit inferior.

Modified Asymptotes - if $\limsup\limits_{x\to a}f(x)-g(x)=\liminf\limits_{x\to a}f(x)-g(x)=0$, then $f(x)$ is asymptotic to $g(x)$ as $x\to a$ and vice versa regardless of interceptions.

IMO, I have never seen something marked as non-asymptotic because of interceptions. If you wanted, you could still squeeze theorem the asymptote using limit superior and limit inferior, so it shouldn't be any problem. Also note that classifying it as non-asymptotic is probably less helpful than classifying it as an asymptote.

EDIT

Also, while you are dealing with the more geometric asymptotes, there are also asymptotes that are more "functional" (as I would put it) in the fields of "Asymptotic Analysis", "Asymptotic Theory", along with the concept of "Big 'O' Notation". Such things allow different meanings to "asymptote", for example:

$$\lim_{x\to\infty}\frac{f(x)}{g(x)}=1\implies f(x)\sim g(x)$$

A famous example of when the above does not agree with our original definition of asymptote is "Stirling's Approximation:

$$\lim_{n\to\infty}\frac{\sqrt{2\pi n}\left(\frac ne\right)^n}{n!}=1,\lim_{n\to\infty}\sqrt{2\pi n}\left(\frac ne\right)^n-n!=-\infty$$

The idea can be thought of in the much more simple example of comparing $f(x)=10^x+x$ and $g(x)=10^x$. Clearly $f(6)=10\text{ Meg}+6$, but $g(6)=10\text{ Meg}$ (Meg=million), but what is $6$ in comparison to $10\text{ Meg}$? It is negligible. But our original definition of asymptote does not consider the $6$ as negligible in comparison to the ten million...

Big $O$ notation is also useful, for example,

$$\frac{x^5+7}{x^2-2}=x^3+O(x^2)$$

all this is really saying is that the rational function has a dominant $x^3$ growth/behavior, and the rest of the behavior is of $x^2$ order or less. This can quickly help determine the behavior of extremely complicated functions by extracting the dominant behavior of a function for either limits or approximations, etc.

• That takes care of the second example. Much appreciated. Is a straight line its own asymptote? if $f(x)-g(x)\to0$ as $x\to a$, is great, but if it's always zero? – JTP - Apologise to Monica Sep 16 '16 at 23:35
• @JoeTaxpayer Hm, I'm not really sure. I mean, the whole idea of asymptotes is that $f(x)\ne g(x)$, but one can treat it as $f(x)\approx g(x)$ for approximations and use of things like squeeze theorem. So I guess it really depends on the context of whether or not $f(x)=g(x)$ is allowed. I mean, it doesn't "approach" $0$, but it is most certainly true that limit is "equal" to $0$... – Simply Beautiful Art Sep 16 '16 at 23:52
• Right. In math we expect precision, and for a definition to be complete and unambiguous. I'm ok with convention changing. When I was in school, a 7 sided figure was a septagon. At 51, I found its called a heptagon now. Just an example. – JTP - Apologise to Monica Sep 17 '16 at 0:45
• @JoeTaxpayer: Every function is asymptotic to itself, yes. (In other words, "asymptotic" is a strictly weaker condition than "equal".) A definition of "asymptotic" that involves exceptions for functions that intersect or are eventually equal to their asymptotes would be completely at odds with standard mathematical usage. – Daniel Hast Sep 17 '16 at 4:04
• One thing to keep in mind is how the analytic notion in your edit differs from the geometrical notion of slant asymptotes, which is used in curve sketching. For example, $f(x)=\frac{x^2+1}{x-1}\sim x$ but the geometrical slant asymptote of $y=f(x)$ is $y=x+1$. Also all linear functions of the same slope are asymptotic to one another, but they have differing slant asymptotes, since a line is its own linear asymptote. – user52817 Sep 18 '16 at 2:48

Mathematically a horizontal line is asymptotic to itself. So for example the horizontal asymptote of the constant function $f(x)=3$ is the line $y=3$.

This issue of whether the definition of a horizontal asymptote should place some sort of restriction on the graph of $y=f(x)$ crossing a horizontal asymptote seems to come up frequently. As a mathematician this seems peculiar, and has always puzzled me. I suspect the root of this resides in the etymology of the word "asymptote" which traces back to something like "not together." Thinking about the etymology in this way might tempt one to think that the mathematical definition of an asymptote perhaps should include some restriction to rule out the notion of a line being asymptotic to itself. But it should not and does not.

What we have here is just a peculiar and interesting drift over time from the etymological meaning of a word to a meaning and usage that is quite different.

• I think a more likely cause is the use of the word "approaches" (as in the OP's highlighted definition) to commonly indicate "limit". We might possibly consider a constant function to have a "degenerate limit", which has the same confusion as other types of degeneracy (triangles, conics, etc.). As mathematicians we like a definition to encompass many things (because proofs are thus more concise), but in daily life people want to distinguish those with separate mental objects. – Daniel R. Collins Sep 17 '16 at 5:39
• @Daniel Collins: A reason why the etymology explanation might be equally likely is found in the history of the parallel postulate beginning with Proclus and leading to the Playfair version of the parallel postulate. In this realm, asymptotic lines are lines that do not meet. In what would eventually emerge as the hyperbolic plane, asymptotic lines are divided into limiting parallels and ultra parallels. What some of these authors termed as an "asymptotic line" was termed by Heath as a "non-secant line." – user52817 Sep 18 '16 at 15:35
• Historically true, but surely that's not a meme in most of our students' minds. – Daniel R. Collins Sep 18 '16 at 17:04

I don't think that a horizontal line being asymptotic to itself is a "trick question we just never really run into" (as you suggest in the comments). Probability theory gives a natural context in which this sort of thing occurs (albeit in a way which is more than just a horizontal line). A standard result is that a continuous function $F$ defined on $(-\infty, \infty)$ is a cumulative distribution function of a continuous random variable if and only if it is monotone increasing and is asymptotic to $0$ as it approaches $-\infty$ and $1$ as it approaches $\infty$. For some distributions (e.g. a normal distribution) these are asymptotes as you typically think of them -- you approach $0$ or $1$ but never get there for any finite value. On the other hand, if the distribution is for a random variable with bounded support then the asymptotes $0$ and $1$ will be achieved long before $\pm \infty$. For example, the graph of the cdf of the standard uniform variable on $[0,1]$ looks like Most of the graph consists of horizontal lines which are equal to the asymptotic values. I don't see any reason to consider this a counterexample to the claim that cumulative distribution functions are always asymptotic to $0$ and $1$.