Continuity over an interval | Limits and continuity | AP Calculus AB | Khan Academy

What we're going to do in this video is explore continuity over an interval. But to do that, let's refresh our memory about continuity at a point. So we say that ( f ) is continuous when ( x ) is equal to ( c ) if and only if, so I'm going to make these two-way arrows right over here: the limit of ( f(x) ) as ( x ) approaches ( c ) is equal to ( f(c) ).

When we first introduced this, we said, "Hey, this looks a little bit technical," but it's actually pretty intuitive. Think about what's happening: the limit as ( x ) approaches ( c ) of ( f(x) ). So let's say that ( f(x) ) as ( x ) approaches ( c ) is approaching some value. If we approach from the left, we're getting to this value. If we approach from the right, we're getting this value. Well, in order for the function to be continuous, I had to draw this function without picking up my pen. Well, the value of the function at that point should be the same as the limit.

This is really just a more rigorous way of describing this notion of not having to pick up your pencil; this notion of connectedness that you don't have any jumps or any discontinuities of any kind. So with that out of the way, let's discuss continuity over intervals. Let me delete this really fast so I have space to work with.

So we say, I'm going to first talk about an open interval, and then we're going to talk about a closed interval because a closed interval gets a little bit more involved. So we say ( f ) is continuous over an open interval from ( A ) to ( B ). So the parentheses instead of brackets; this shows that we're not including the end point. So this would be all of the points between ( x = A ) and ( x = B ) but not equaling ( x = A ) and ( x = B ). So ( f ) is continuous over this open interval if and only if, if and only if, ( f ) is continuous over every point in the interval.

So let's do a couple of examples of that. So let's say we're talking about the open interval from -7 to -5. Is ( f ) continuous over that interval? Let's see, we're going from -7 to -5, and there's a couple of ways you could do it. There's the not so mathematically rigorous way where you could say, "Hey look, if I start here, I can get all the way to -5 without having to pick up my pencil." If you wanted to do it more rigorously and you actually had the definition of the function, you might be able to do a proof that for any of these points over the interval, the limit as ( x ) approaches any one of these points of ( f(x) ) is equal to the value of the function at that point.

It's harder to do when you only have a graph. When you only have a graph, you can only just do it by inspection and say, "Okay, I can go from that point to that point without picking up my pencil," so I feel pretty good about it. Now let's do another interval. Let's say the, so let me put a check mark here that it is continuous.

Let's think about the interval from -2 to positive 1, the open interval. So this is interesting because the function at -2 is up here. If you really wanted to start at -2, you would have to start here and then jump immediately down as soon as you get slightly larger than -2 and then keep going. But this is an open interval, so we're not actually concerned with what exactly happens at -2. We're concerned with what happens when we are all the numbers larger than -2.

So we would actually start right over here and then we would go to 1, and once again, based on the intuitive "I didn't have to pick up my pen" idea, this function would be continuous over this interval. So what's an example of an interval where the function would not be continuous? Well, think about the interval from, well this is a pretty straightforward one, the open interval from 3 to 5. The function is here when ( x ) is equal to 3, but if we wanted to get to 5, it looks like we're asymptoting. It looks like we're asymptoting up towards infinity, and we just keep on going for a very long time and then we would have to pick up our pencil and jump over, and then we would come back down right over here.

So here we are not continuous over that interval. Now let's think about the slightly more involved case when you have a closed interval. ( f ) is continuous over the closed interval from ( A ) to ( B ). So this includes not just the points between ( A ) and ( B ), but the end points as well. If and only if ( f ) is continuous over the open interval, and the one-sided limits—let me write this— and the limit as ( x ) approaches ( A ) from the right of ( f(x) ) is equal to ( f(A) ), and the limit as ( x ) approaches ( B ) from the left of ( f(x) ) is equal to ( f(B) ).

Now what's going on here? Well, it's just saying that the one-sided limit when you're operating within the interval has to approach the same value as the function. So for example, if we said the closed interval from -7 to -5, well this one is still reasonable; you know just based on the picking up your pencil thing, you don't have to pick up your pencil.

What you would do is at the end point and at -7, this function is just plain old continuous. But if it wasn't defined over here, it could still be continuous because you'd do the right-handed limit towards it and you'd say, "Okay, the right-handed limit is equal to the value of the function." And then at this end point, at the second end point, you'd say, "Okay, the left-handed limit is equal to the function," even if it wasn't defined here, even if the two-sided limit were not defined.

We could actually look at an example of that. If we were looking at the interval from the closed interval from -3 to -2, notice I did not have to pick up my pencil. I'm including -3, and I'm getting all the way to -2. If you knew the analytic definition of this function, you could prove that, "Hey, the limit at any of these points inside between -3 and -2 is equal to the value of the function."

The function is clearly at -3, the function is just plain old continuous. The two-sided limit approaches the value of the function, but at -2, the two-sided limit does not exist. When you approach from the left, it looks like you're approaching 0; ( f(x) ) is equal to 0. When you approach from the right, it looks like ( f(x) ) is approaching -3.

So even though the two-sided limit does not exist, we can still be good because the left-handed limit does exist and the left-handed limit is approaching the value of the function. So we actually are continuous over that interval. But then if we did the closed interval from -2 to 1, pause the video and think about based on what we just talked about: are we continuous over this interval?

Well, we're going from -2 to 1, and -2 is the lower bound. So is this right over here true? Is the limit as we approach -2 from the right the same thing as ( f(-2) )? Well, the limit as we approach from the right seems to be approaching -3, and ( f(-2) ) is 0. So this limit does not, this does not hold; the limit as we approach from the right and the value of the function are not the same.

So we do not have that, I guess you could say that one-sided continuity at -2, and that also makes sense. If I start at -2, let me do this in a color you can see. If I start at -2 and I want to go the rest of the interval to 1, I have to pick up my pencil, pick up my pencil, go here, and then keep on going. So this is, we are not continuous over that interval.