# Discontinuity points challengeÂ 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.