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## AP®︎/College Calculus AB

### Unit 1: Lesson 10

Exploring types of discontinuities# Types of discontinuities

AP.CALC:

LIM‑2 (EU)

, LIM‑2.A (LO)

, LIM‑2.A.1 (EK)

A function being continuous at a point means that the two-sided limit at that point exists and is equal to the function's value. Point/removable discontinuity is when the two-sided limit exists, but isn't equal to the function's value. Jump discontinuity is when the two-sided limit doesn't exist because the one-sided limits aren't equal. Asymptotic/infinite discontinuity is when the two-sided limit doesn't exist because it's unbounded.

## Want to join the conversation?

- Is an asymptotic discontinuity any different than an infinite discontinuity?(24 votes)
- They are the same thing – if you look in the
`About`

text it actually says "Asymptotic/infinite discontinuity".(23 votes)

- I understand that classification of discontinuities is 3 types

i. point removable

ii. jump discontinuities

iii.asympotic discontinuities

anything also(4 votes)- There are also oscillating discontinuities. Look at the graph of f(x)=sin(1/x). It has no value or limit at x=0.(5 votes)

- Hi, I am learning how to evaluate functions by direct substitution right now. I was wondering why simply substituting or re-arranging a function would automatically give us the limit at that point. For example, in the two graphs on the left in this video, the y-value is defined at the x-value but the limit either doesn't equal that same y-value or doesn't exist. I want to see the actual functions that could result in these two graphs to better understand why we can directly substitute without fear of scenarios like these two. Is it only possible for piece-wise functions to create these types of scenarios? I feel like I am overlooking something and would really appreciate the help. Thanks in advance!(4 votes)
- A function can be determined by direct substitution if and only if lim_(x->c)_ f(x) = f(c). In other words, as long as the function is not discontinuous, you can find the limit by direct substitution.

There is also another way to find the limit at another point, and that is by looking for a determinant for the indeterminate form by using other methods and defining it by using another function. For example, lim_(x->2) (x^2 + 4 x - 12)/(x - 2), determined directly, equals (0/0), indeterminant form. However, there are many ways to determine a function by simply simplifying the function when direct substitution yields the indeterminant form. For this example, you could simply factor the limit to get lim_(x->2)_ (x+6), x ≠ 2. The constraint is added to be mathematically correct when it comes to being equivalent to the limit beforehand. However, say you found a function that is similar to the simplified function, only without the constraint, called g(x) = (x+6). You can define that as your new limit: lim_(x->2) g(x) = 8, thus lim_(x->2) (x^2 + 4 x - 12)/(x - 2) = 8.

If you want to learn more, go to this page to see some more situations in which it's possible to do a direct substitution: https://www.khanacademy.org/math/ap-calculus-ab/ab-limits-new/ab-1-7/a/limit-strategies-flow-chart(1 vote)

- Is there any difference between saying that a limit doesn't exist and saying that a limit is unbounded? I've only ever heard Sal saying a limit doesn't exist/there is no limit when a limit is being taken from both sides. This is the first time I can remember Sal saying something similar regarding a limit being taken from one side, however he called the limit unbounded this time.

I hope I explained myself clearly and if anyone could answer this question, that would be appreciated, thanks!(3 votes) - Is a quadractic formula discontinous(1 vote)
- Well, the quadratic formula is a formula, so it can't be graphed.

However, a function related to the quadratic formula, a quadratic polynomial, is continuous over its entire interval.(4 votes)

- I can't understand why the value of the y=x^2 graph at x=3 is 4, and not 9. Probably an obvious answer, but it's eluding me!(2 votes)
- So can you see the dot that is separated from the curve?(2 votes)

- What about the function which has one sided limit? For example if lim_(x->0) when approaching
**from the right**exists, but lim_(x->0) when approaching**from the left**is*asymptote*?

Does this qualify as asymptotic discontinuity or some kind of mix of jump/asymptotic discontinuity?(2 votes)- That would be an asymptotic discontinuity. Asymptotic discontinuities are defined as occurring when at least one of the one-sided limits are undefined.(2 votes)

- Do sharp peaks or turns in the line, such as the ones in some parametric equations, count as a discontinuity?(1 vote)
- The derivative of said functions would be discontinuous, but as long as the line never breaks it is continuous. Still, sharp turns or other sudden changes in slope will make the function non differentiable. So still something you have to keep an eye out for.(4 votes)

- Is it still a jump discontinuity if the hole and the next point don't share the same x value?(2 votes)
- What would be an example of a function with both a removable and a non-removable discontinuity?(2 votes)

## Video transcript

- [Instructor] What we're
going to do in this video is talk about the various
types of discontinuities that you've probably seen
when you took algebra, or precalculus, but then
relate it to our understanding of both two-sided limits
and one-sided limits. So let's first review the
classification of discontinuities. So here on the left,
you see that this curve looks just like y equals x squared, until we get to x equals three. And instead of it being three squared, at this point you have this opening, and instead the function at
three is defined at four. But then it keeps going
and it looks just like y equals x squared. This is known as a point, or a removable, discontinuity. And it's called that for obvious reasons. You're discontinuous at that point. You might imagine defining
or redefining the function at that point so it is continuous, so that this discontinuity is removable. But then how does this
relate to our definition of continuity? Well, let's remind ourselves
our definition of continuity. We say f is continuous, continuous, if and only if, or let me write f continuous at x equals c, if and only if the limit as x approaches c of f of x is equal to the
actual value of the function when x is equal to c. So why does this one fail? Well, the two-sided limit actually exists. You could find, if we say
c in this case is three, the limit as x approaches three of f of x, it looks like, and if you
graphically inspect this, and I actually know this is the
graph of y equals x squared, except at that discontinuity
right over there, this is equal to nine. But the issue is, the way
this graph has been depicted, this is not the same thing
as the value of the function. This function f of three, the way it's been graphed, f of three is equal to four. So this is a situation where
this two-sided limit exists, but it's not equal to the
value of that function. You might see other
circumstances where the function isn't even defined there, so that isn't even there. And so, once again, the limit might exist, but the function might
not be defined there. So, in either case, you aren't
going to meet this criteria for continuity. And so that's how a point
or removable discontinuity, why it is discontinuous with regards to our limit
definition of continuity. So now let's look at this second example. If we looked at our
intuitive continuity test, if we would just try to trace this thing, we see that once we get to x equals two, I have to pick up my
pencil to keep tracing it. And so that's a pretty good
sign that we are discontinuous. We see that over here as well. If I'm tracing this function,
I gotta pick up my pencil to, I can't go to that point. I have to jump down here, and then keep going right over there. So in either case I have
to pick up my pencil. And so, intuitively, it is discontinuous. But this particular type of discontinuity, where I am making a jump from one point, and then I'm making a jump
down here to continue, it is intuitively called a jump discontinuity, discontinuity. And this is, of course, a
point removable discontinuity. And so how does this relate to limits? Well, here, the left and
right-handed limits exist, but they're not the same thing, so you don't have a two-sided limit. So, for example, for
this one in particular, for all the x-values up to
and including x equals two, this is the graph of y equals x squared. And then for x greater than two, it's the graph of square root of x. So in this scenario, if you were to take the limit of f of x as x approaches two from the left, from the left, this is going to be equal to four, you're approaching this value. And that actually is the
value of the function. But if you were to take the
limit as x approaches two from the right of f of x, what is that going to be equal to? Well, approaching from the right, this is actually the square root of x, so it's approaching
the square root of two. You wouldn't know it's
the square root of two just by looking at this. I know that, just because when I, when I went on to Desmos
and defined the function, that's the function that I used. But it's clear even visually that you're approaching
two different values when you approach from the left than when you approach from the right. So even though the one-sided limits exist, they're not approaching the same thing, so the two-sided limit doesn't exist. And if the two-sided limit doesn't exist, it for sure cannot be equal to the value of the function there, even
if the function is defined. So that's why the jump
discontinuity is failing this test. Now, once again, it's intuitive. You're seeing that, hey, I gotta jump, I gotta pick up my pencil. These two things are not
connected to each other. Finally, what you see here is, when you learned precalculus, often known as an
asymptotic discontinuity, asymptotic, asymptotic discontinuity, discontinuity. And, intuitively, you
have an asymptote here. It's a vertical asymptote at x equals two. If I were to try to trace the graph from the left, I would just keep on going. In fact, I would be doing
it forever, 'cause it's, it would be infinitely, it would be unbounded as
I get closer and closer to x equals two from the left. And if try to get to x
equals two from the right, once again I get unbounded up. But even if I could, and when I say it's unbounded,
it goes to infinity, so it's actually impossible in a mortal's lifespan to
try to trace the whole thing. But you get the sense that,
hey, there's no way that I could draw from here to here
without picking up my pencil. And if you wanna relate it
to our notion of limits, it's that both the left and right-handed
limits are unbounded, so they officially don't exist. So if they don't exist, then
we can't meet these conditions. So if I were to say, the limit as x approaches two from the
left-hand side of f of x, we can see that it goes unbounded
in the negative direction. You might sometimes see someone
write something like this, negative infinity. But that's a little
handwavy with the math. The more correct way to say
it is it's just unbounded, unbounded. And, likewise, if we
thought about the limit as x approaches two from the right of f of x, it is now unbounded
towards positive infinity. So this, once again, this is also, this is also unbounded. And because it's unbounded and
this limit does not exist, it can't meet these conditions. And so we are going to be discontinuous. So this is a point or
removable discontinuity, jump discontinuity, I'm jumping, and then we have these
asymptotes, a vertical asymptote. This is an asymptotic discontinuity.