14
$\begingroup$

This is a general question. A function is said to be continuous. Can it still have vertical asymptotes? Looking at the definition of continuity, I would say no. Because near a vertical asymptote x-delta might have an y of close to minus infinity, while x+delta might have a value of near +infinity, for example.

$\endgroup$
  • 1
    $\begingroup$ Continuous over the domain of all real numbers or continuous over some open interval(s)? $\endgroup$ – Ahmed S. Attaalla Aug 14 '16 at 20:10
  • 6
    $\begingroup$ If it has an assymptote at $x = a$ the function might be continuous on all points $x \ne a$. As it is undefined at $x = a$, $a$ is not in the domain of $f$. So $f$ may be continuous on all points of its domain. But it isn't continuous on all points of R because it isn't defined on all points of R. But "continuous" in higher math does not mean continuous on all points of R. It means continuous on all points of its domain. So yes it can be continuous. $\endgroup$ – fleablood Aug 14 '16 at 20:26
  • $\begingroup$ Your reasoning isn't valid as your delta would be too big. The def of continuous at x means there exists a delta. Not all deltas. So that is simply not the correct delta. But there can be a delta that is smaller. Take an assymptote at $a$. Take point x. If delta < |x - a| then NO x+delta and x - delta do not cross the assymptote. $\endgroup$ – fleablood Aug 14 '16 at 20:30
  • $\begingroup$ The domain is not specified in the exercise. Nor is it specified if it is continuous over R. So I guess it is left open in my exercise, which is ok for the questions I have to answer. At least now I understand why there is a question mark in the answer book instead of an asymptote at certain values for x. tt's because you can't tell for sure what is the case. $\endgroup$ – babipsylon Aug 14 '16 at 20:38
  • $\begingroup$ In high school, pre-calculus "continuous" is sometimes defined to be required to be defined for all x in R. This is NOT the case for calculs and higher. We only consider where x where f(x) is defined. In other words the domain of f. It doesn't matter what the domain of f is exactly. If f has a vertical assymptote at $a$ then $a$ is not in the domain and we do not consider $a$ at all in determining if $f$ is continuous. $f$ may have as many vertical assymptotes as we wish and still be continuous. $\endgroup$ – fleablood Aug 14 '16 at 21:09
29
$\begingroup$

It doesn't make sense to ask whether a function is continuous at a point where it's not defined. So if it has a vertical asymptote, then that's not a point of discontinuity, but rather a point that is not part of the domain.

To make it clear what I mean by "doesn't make sense", the definition of continuity at a point $x$ involves the expression $|f(x) - f(y)|$. If $x$ is such that $f(x)$ doesn't make sense, then neither does that expression, so asking about continuity is meaningless.

$\endgroup$
  • 7
    $\begingroup$ On the other hand, it does make sense to ask whether a function "could be made continuous" at a point where it's not defined.For example, the sinc function (en.wikipedia.org/wiki/Sinc_function) is defined as $sinc(x) = sin(x)/x$ for $x \ne 0$ and it doesn't make sense to ask whether that is continuous at $x = 0$; but it approaches $1$ as $x$ approaches $0$, so we can just define $sinc(0) = 1$ and the resulting function is continuous at $x = 0$. $\endgroup$ – Robin Saunders Aug 14 '16 at 22:46
  • 2
    $\begingroup$ Some definitions of "vertical asymptote" speak only with the infinite limit, so that a function like $f(x)=1/x$ completed by $f(0)=0$ has a vertical asympote - and your argument would not apply education.ti.com/html/t3_free_courses/calculus84_online/mod07/… $\endgroup$ – leonbloy Aug 15 '16 at 1:20
  • $\begingroup$ Glad to see people are catching up with knowing that it doesn't make sense. I remember a time where people would claim that such a function would be non-continuous at these points, as if it was something that carried any meaning. $\endgroup$ – Git Gud Aug 15 '16 at 7:38
  • $\begingroup$ @GitGud Back when I took introductory calculus, one of the problems on a mandatory assignment was "Show that $1/x$ is continuous." I don't know how many times the lecturer and TA's had to say "... where it is defined" to people (including me) asking whatever they could mean asking something like that. $\endgroup$ – Arthur Aug 15 '16 at 7:58
  • 2
    $\begingroup$ @GitGud: If your framework of thinking is measure theory and measures which are absolutely continuous with respect to Lebesgue measure, the statement does make sense. Your functions are actually equivalence classes of functions that differ on sets of measure zero, and the statement that "$f$ is discontinuous at $x$" actually means there's no member of $f$'s equivalence class that's continuous at $x$. $\endgroup$ – R.. Aug 15 '16 at 16:15
12
$\begingroup$

The function $f\colon \Bbb R\setminus\{0\}\to\Bbb R$ given by $f(x)=\frac 1x$ is continuous on all of its domain. Yet there is a vertical asymptote.

$\endgroup$
3
$\begingroup$

An important notion not mentioned in any answer so far is compactness. There is a formal definition of this (https://en.wikipedia.org/wiki/Compact_space) which can be applied to any abstract domain that your function might be defined on, but if your domain is a subset of the number line then it's enough to check whether it's bounded (doesn't "stretch off" to infinity) and closed (includes all its "end points"). Other answers have given various examples of functions which were continuous on their domain and still included a vertical asymptote, but that asymptote was always an "end point" of the domain that was not included in the domain itself; so the domain was not closed and hence not compact.

In general, the image of a compact region under any continuous function will always be compact; so, if a function has values in the numberline (or any other metric space) then the set of values of the function on any compact region of its domain must be compact, and hence bounded, and there can be no vertical asymptote.

$\endgroup$
1
$\begingroup$

Yes. $1/x$ is continous on (0,1) and has a vertical asymptote at 0.

$\endgroup$
  • $\begingroup$ what about $[0,1]$ ? $\endgroup$ – DeepSea Aug 14 '16 at 20:12
  • 2
    $\begingroup$ @DeepSea It's not defined on $0$, so no. $\endgroup$ – Arthur Aug 14 '16 at 20:12
  • $\begingroup$ Yes. A continuous function can't have a vertical asymptote over a point in the domain at which it is defined. $\endgroup$ – Thompson Aug 14 '16 at 20:15
1
$\begingroup$

The standard definition of continuity only considers points in the domain of the function. Note that by common understanding, a point where a function is undefined, like a vertical asymptote, is not included in its domain. Therefore, a function can have a vertical asymptote and still be a continuous function.

For example, from Stein and Barcellos, Calculus and Analytic Geometry, 5th Edition (sec. 2.8):

Definition Continuous function. Let f be a function whose domain is the x axis or is made up of open intervals. Then f is a continuous function if it is continuous at each number a in its domain.

EXAMPLE 1 Use the definition of continuity to decide whether $f(x) = 1/x$ is continuous.

SOLUTION [reasoning about definition]... Thus $1/x$ is continuous at every number in its domain. Hence it is a continuous function.

Even though, of course, $1/x$ has a vertical asymptote at $x = 0$.

$\endgroup$
1
$\begingroup$

"Because near a vertical asymptote x-delta might have an y of close to minus infinity, while x+delta might have a value of near +infinity, for example."

Then you just have to choose a smaller delta.

Take $f(x) = 1/x$ for example. It is continuous on all $x \ne 0$. And as $f(0)$ is undefined, it is continuous on all points in its domain.

Pick a point $x_1 > 0$. Then the $\delta$ you choose must be $\delta < |x_1 - 0|=x_1$. But that is always possible. If $\delta < x_1$ then $0 < x_1 - \delta < x_1 < x_1 + \delta$ and if $|x - y| < \delta$ then $0< \frac 1{x+ \delta} < \frac 1y < \frac 1{x-\delta}$ does not have the problem of containing a range of unbounded values.

The same argument holds for $x_2 < 0$.

=====

Practical example: Let $x_1 = 1/\text{googol} = 10^{-100}$. Is $f(x) = 1/x$ continuous at $x= x_1$? $x_1$ is pretty damned close the the assymptote, isn't it.

Let $\epsilon > 0$. We want to find a $\delta > 0$ so the for all $y$ such that $|y- 10^{-100}| < \delta$ then $|f(y) - f(10^{-100})| = |1/y - 10^{100}| < \epsilon$.

To find such I need $-\epsilon < 1/y - 10^{100} < \epsilon$ or $10^{100} - \epsilon < 1/y < 10^{100} + \epsilon$

or $\frac 1{10^{100}+\epsilon} < y < \frac 1{10^{100} - \epsilon}$.

$\frac 1{10^{100} +\epsilon} -x_1 < y-x_1 > \frac 1{10^{100} - \epsilon}-x_1$

So for $\delta = \min (|10^{100} - \frac 1{10^{100} +\epsilon},|\frac 1{10^{100} -\epsilon} - 10^{100}|)$, as long as $|x_1 - y| < \delta$ then $|1/x_1 - 1/y| < \epsilon$.

Note: $0 < \delta < 1/10^{100}$

So $f$ is continuous at $x = 1/\text{googol}$. But the delta we had to find was ever smaller than $1/\text{googol}$.

$\endgroup$
  • $\begingroup$ δ>|x1−0| I guess. But I guess the absolute value property of the definition just doesn't apply if there is a point not part of the domain. $\endgroup$ – babipsylon Aug 14 '16 at 20:58
  • $\begingroup$ continuous means for every $x_1$ and every $[f(x_1) - e, f(x_1)+e]$ we can find a $[x_1 -d, x_1 + d]$ so that $f([x_1 - d, x_1 + d]) \subset [f(x_1) - e, f(x_1)+e]$. It doesn't matter if there are points "missing" from $[x_1 -d, x_1 +d]$ as long as $f([x_1 -d, x_1 + d] )\subset [f(x_1) -e, f(x_1)+e]$. But if there is an assymptote we can't have the point in $[x_1 - d, x_1+d]]$ because $f([x_1 -d, x_1 + d])$ "blows up". Absolute value is irrelevent. It just makes everything measured as a positive distance. $\endgroup$ – fleablood Aug 14 '16 at 23:19
  • $\begingroup$ ...but it doesn't matter if there are asymptotes because we can always find smaller deltas so that the points of assymptotes are not in $[x_1 -d, x_1 + d]$. If there is a huge honking assymptote at x=a, it doesn't matter. For $x_1 \ne a$ we can find $f([x_1 -d, x_2 + d]) \subset [f(x_1) - e, f(x_1) + e]$ so long as $x_1-d$ and $x_2 + d$ are both on the same side of $a$. Nothing to do with absolute values. $\endgroup$ – fleablood Aug 14 '16 at 23:24
-2
$\begingroup$

How about $f(x) = x^{1/3}$ (real branch)? The asymptote in $(0,0)$ is vertical and the function is continuous on the reals.

$\endgroup$
  • 1
    $\begingroup$ That's not a vertical asymptote, that's a vertical tangent line. $\endgroup$ – Noah Schweber Aug 15 '16 at 19:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.