Studying for a test? Prepare with these 21 lessons on Limits and continuity.
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# Discontinuity points challenge example

Video transcript
The graph of a function f is shown below. If both the limit of f of x as x approaches k and f of k exist, and f is not continuous at k, then what is the value of k? So we have to find a k where f is not continuous at k, but f of k is defined. And the limit as x approaches k of f of x is also defined. So the easiest ones to spot out just with our eyes might just to be to see where f is not continuous, where f is not continuous. So you see here when x is equal to negative 2, the function is not continuous. It jumps from up here. It looks like it's approaching 3. And then it jumps down to negative 3. So this is one of our candidates. This is one of our candidates for k. The other discontinuity happens when x is equal to 3. Once again, we jump down from-- looks like 4 and a 1/2 all the way to negative 4. So that's another candidate. That's a point where f is not continuous. And then we have when x is equal to 8, we have this. As we approach this, it looks like we're getting to 1. But then we jump right at x equals 8. And then we continue from 1 again. So this is our other candidate. So these are the three candidates where the function is not continuous. Now let's think about which of these points, which of these x values, does f of k exist. So if one of these is k, does f of k exist? Well f of negative 2 exists. f of 3 exists, right over here. That's f of 3. This is f of negative 2. And f of 8, all exist. So all of these potential k's meet this constraint-- f of k exists, and f is not continuous at k. So that's true for x equals 8, 3, or negative 2. Now let's look at this first constraint. The limit of f of x as x approaches k needs to exist. Well, if we tried to look at x equals 2, the limit of f of x as x approaches negative 2-- not 2, as x approaches negative 2 here-- the limit from the left, the limit from values lower than negative 2, it looks like our function is approaching something a little higher. It looks like it's a little higher than 3. And the limit from the right, it looks like our function is approaching negative 3. So this one, the limit does not exist. You get a different limit from the left and from the right. Same thing for x equals positive 3. The limit from the left seems like it's approaching 4 and 1/2, while the limit from the right looks like it's approaching negative 4. So this is also not a candidate. So we only have one left. So for this one, the limit should exist. And we see the limit as f of x as x approaches 8 from the negative direction, it looks like f of x is approaching 1. And it looks like, as we approach 8 from the positive direction, the limit of f of x as x approaches 8 from the positive direction. It's also equal to 1. So your left- and your right-sided limits approach the same value. So the limit of f of x as x approaches 8 is equal to 1. This limit exists. Now, the reason why the function isn't continuous there is that the limit of f of x as x approaches 8, which is equal to 1, it does not equal the value of f of 8. f of 8, we're seeing, is equal to 7. So that's why it meets the last constraint. The function is not continuous there. The function exists. It's defined, f of 8 is equal to 7. And the limit exists. But the limit of f of x as x approaches k is not the same thing, or is not the same as the value of the function evaluated at that point. And so x equals 8 meets all of our constraints. So we could say k is equal to 8.