 This problem now is going to look at continuity of a function over an interval. So no longer are we considering just one single x value, but rather a span of x values. What often will make the difference here is whether it is an open or closed interval and what is going on in each of those endpoints in terms of that open versus closed interval. So the functioning question here is x over x minus 2 and I have a graph of it over here on the next page. You'll notice that at x equals 2 we do have a vertical asymptote. And the other points you can simply generate either off of your graphing calculator or by substituting x values in as I had done. So the first one we're going to look at is the interval from negative infinity up to and including 0. So starting at the far left coming along that branch in the second quadrant up to and including 0. All is good there. There's no breaks. There's no holes. There's no asymptotes. So this function would be continuous and I'm just going to for efficiency sake see for continuous. Now notice the next interval is from 0 including 0 up to through positive infinity. So basically x greater than or equal to 0. So now we're going to be starting here at 0 at the origin and going off to the right. Well the asymptote gets in the way as you do that because of course this function is not continuous there. So we would say in this case that this function is discontinuous on that particular interval. Now the next interval we're going to look at is the purely open interval from 0 to 2. So we're looking at from 0 now it's not including 0 up through 2 but not including 2. That's the important part here. In that case this is continuous on that particular interval because we are not including 2 in consideration. It's getting really close to 2 but it's not hitting 2. So in this case that one is going to be continuous. But look at what happens on the next interval. Now we are open on 0 up to and including 2 but that's where our asymptote is. So now that we're including 2 in consideration because of the asymptote we would have to say this particular interval though is discontinuous. Alright so that's where I was talking about the issue really being this whether it's an open or a closed interval on a particular endpoint. Something similar is going to happen with the next two intervals we are going to consider. If we consider the interval that is closed on 2 through infinity while we're starting where the asymptote is and we are including that value of 2. That's what poses the problem because there's the asymptote there. So we would have to say this particular interval is discontinuous. But in the next one, our last one if we think of 2 through infinity but not including 2 where the asymptote is really we're just talking about this particular branch right here in the first quadrant. In which case then we're okay and we would say that is continuous.