The difference between a "removable discontinuity" and a "vertical asymptote" is that we have a R. discontinuity if the term that makes the denominator of a rational function equal zero for x = a cancels out under the assumption that x is not equal to a.

Othewise, if we can't "cancel" it out, it's a vertical asymptote.

Could there potentially be an intuitive way to understand this? Why is this the case?

In case of an R. discontinuity, I figure that it's a single isolated point (x = a, in this case) that the function is not defined at but otherwise continuous.... but isn't that also the case for the function with the verticla asymptote? Isn't this also just not defined for that isolated point, but otherwise continuous? So why do they behave so differently, with a function with a R.D just being a regular function but with a "circle" over the undefined point, while a vertical asymptote makes the function do some crazy stuff (i.e. jump off towards infinity when getting closer to a).

  • 1
    $\begingroup$ The key distinction between a removable discontinuity and a discontinuity which corresponds to a vertical asymptote is that $\lim_{x\to a}f(x)$ exists in the case of a removable discontinuity, but $\lim_{x\to a^{+}}f(x)$ or $\lim_{x\to a^{-}}f(x)$ is infinite in the case of a vertical asymptote. $\endgroup$
    – user84413
    Commented Aug 27, 2014 at 18:53

3 Answers 3


The patients have been skiing and they all have a fracture.

Patient A) f(x) = 5 if x is not 0, and 0 is not in dom(f).

Patient B) f(x) = 5 if x is not 0 and f(0)=17

Patient C) f(x) = |x|/x

Patient D) f(x) = 1/x

Patients A and B have a removable discontinuity at 0, because the left and right hand limits are both 5. (They exist and are equal.) Patient A has lost one point of bone and we no longer have it, but with the miracle of modern medicine it is easily replaced. Patient B) has the point, but it's way off at 17 when it should be at 5. Both operations are simple, and the patients will be back on the slopes soon.

You have shown a callous disregard for patient C), whose fracture is indeed more serious, because it is a jump discontinuity. At 0, the left-hand limit is -1 and the right-hand limit is 1. (At 0, the left and right hand limits exist, but are not equal.) You can imagine what the x-ray looks like, or just sketch the graph.

Patient D) will never ski again, because there is an infinite discontinuity at 0. Even if just one of the left or right hand limits was infinite (either + or -), the fracture is inoperable.

  • 1
    $\begingroup$ interesting medical analogy (+1) $\endgroup$
    – G Cab
    Commented Jul 14, 2017 at 23:27

This is an answer on the question from the khanacademy.org Full credit for this answer goes to the user "Tronax" from khanacademy.org:

I will try to show the difference between the two types of what we previously called "undefined".

Let's start with the "removable discontinuity", which actually says, that the graph at the point of x could be absolutely any y! And there is no way to find it. It could be 0, -7, 65, 23/7, 3982.3 etc. Math guys agreed between themselves, that if they can't understand it better and find its value - it does not exist. So now we just draw an empty circle on the curve of our graphs in spots that are known to have a removable discontinuity.

This happens if the function is of a form that includes the same non constant factor in both numerator and denominator. For example y = (x+1)(x+2) / (x+1). Notice (x+1) factor is both on the numerator and on the denominator. For x=-1, (x+1) equals zero, so we say that at x=-1 function has a removable discontinuity.

But... why?

Because when you input x=-1 and try to solve for y: y = (-1+1)(-1+2) / (-1+1) y = 0*(-1)/0 y = 0/0 or (this is one of interpretations): 0y = 0 You could substitute any number into y, this expression will fit them all, since anything multiplied by 0 equals 0. This is removable discontinuity. The graph around the point of it, looks just like it would, if there was no removable discontinuity.

The second type is the "vertical asymptote". It occurs when for some x, the denominator (and only denominator) equals zero. It's somewhat easier to understand. Let's think about what happens when we see 8/4. This says: how many times, can we fit the denominator 4 into 8? It takes 2 times. For 8/2 it takes 4 times; for 8/1 - 8 times; 8/0.5 - 16 times; 8/0.25 - 32 times. The lower and closer to zero our denominator gets, the more times we can fit it into the same 8. When it becomes really really small, it can take millions of times! When it gets such a small number that we can barely distinguish it from zero, it seems like it would take almost infinite times of that number to fit our 8! That's why when denominator approaches 0, we say that the result approaches infinity. When x approaches 0, y=8/x approaches infinity and y=-8/x approaches negative infinity.

Don't be confused, when x actually equals zero, y=8/x is undefined! In this case, you could say that even infinity is not enough :)

When saying "approaches zero", we only mean that it gets very very very close to zero, but not zero.

"Vertical asymptote at x = k" only refers to what happens before and after k on the graph. x = k itself is undefined.

The graph before and after vertical asymptote quickly gets almost parallel to the y-axis.

Hope this is helpful.

Source: https://www.khanacademy.org/math/algebra2/rational-expressions-equations-and-functions/discontinuities-of-rational-functions/v/discontinuities-of-rational-functions


Let me try to write a more concise and mathematically precise answer by comparing definitions. The definition for a horizontal asymptote at $y = k$ is:

$$ \lim_{x \rightarrow \infty} f(x) = k \qquad \text{or} \qquad \lim_{x \rightarrow -\infty} f(x) = k $$

In words, as $x$ values get increasingly large or small, the function approaches a $y$ value $k$. This has nothing to do, per se, with the function being defined or not at $k$.

But the definition for a removable discontinuity at $x_k$—notice that this is an $x$ value rather than a $y$ value—the definition is:

$$ \lim_{x \rightarrow x_k} f(x) = L < \infty \quad \text{ and } \quad f(x_k) \neq L $$

Comparing the two definitions in words: for an asymptote, we care about how $f(x)$ behaves as $x$ gets really large or really small, while for a removable discontinuity, we care about how $f(x)$ behaves as $x$ gets really close to some $x$ value $x_k$. Here is a visualization of that difference, with the asymptote on the left and the removable discontinuity on the right:

enter image description here

There are subtleties to this distinction that I am glossing over, but I think this captures the big picture difference.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .