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## Continuity at a point

Current time:0:00Total duration:7:19

# Worked example: Continuity at a point (graphical)

AP Calc: LIM‑2 (EU), LIM‑2.A (LO), LIM‑2.A.2 (EK)

## Video transcript

- [Voiceover] We have the
graph of y is equal to g of x right over here, and what
I wanna do is I wanna check which of these statements
are actually true and then check them off. And like always, I encourage
you to pause the video and see if you can work
through this on your own. So let's look at this first statement. So this first statement says
both the limit of g of x as x approaches six
from the right-hand side and the limit as x approaches
six from the left-hand side of g of x exist. All right, so let's first
think about the limit of g of x as x approaches
six from the right-hand side, as we approach six from
values greater than six. So if we look over
here, we could say okay, when x is equal to nine, and
g of nine is right over there, g of eight is right over here, g of seven is right over here, looks like it's between negative
three and negative four, g of 6.5 looks like it's a little bit, it's still between negative
three and negative four but it's closer to negative three. G of 6.1 is even closer to negative three. G of 6.01 is even closer
to negative three, so it looks like the limit from the right-hand side does exist. So it looks like this one exists. Now let's see, and I'm just
looking at it graphically. That's all they can expect you to do in an exercise like this. Now let's think about the
limit as x approaches six from the left-hand side. So I could start anywhere,
but let's say when x is equal to three, g of three is
a little more than one. G of four looks like there's
a little bit less than two. G of five looks like it's close to three. G of 5.5 looks like it's
between five and six. G of 5.75 looks like
it's approaching nine. And as we get closer and closer, as x gets closer and
closer to six from below, from values to the left of six, it looks like we're unbounded,
we're approaching infinity. And so technically, we would
say this limit does not exist. So this one does not exist. So I won't check this one off. Some people say the limit
is approaching infinity, but that technically is, infinity is not a value that you can say it is approaching in the
classical formal definition of a limit. So for these purposes, we would just say this does not exist. Now let's see, they say the
limit as x approaches six of g of x exists. Well, the only way that the limit exists is if both the left, if
both the left, the left, and the right limits exist and
they approach the same thing. We'll, our limit as x approaches
six from the negative side or from the left-hand
side, I guess I could say, does not even exist. So this cannot be true. So that's not gonna be true. The first one's not gonna be true. G is defined as x equals six. So at x equals six, it doesn't
look like g is defined. Looking at this graph, I can't tell you what g of six should be. We have an open circle
over here, so g of six does not equal to negative three, and this goes up to infinity, and we have a vertical
asymptote actually drawn right over here at x equals six. So g is not defined at x equals six. So I'll rule that one out. G is continuous at x equals six. Well, you can see that
it goes up to infinity then it jumps down, back
down here, then it continues. So when you just think about
it in commonsense language, it looks very discontinuous. And if you wanna think
about it more formally, in order for something to be continuous, the limit needs to exist at that value. The function needs to be
defined at that value, and the value of the
function needs to be equal to the value of the limit and neither of these, the
first two conditions are true and so these can't even equal
each other because neither of these exist. So this is not continuous at x equals six and so the only think I can
check here is none of the above. Let's do another one of these. So the first statement,
both the right-hand and the left-hand limit
exists as x approaches three. So let's think about it. So x equals three is
where we have this little discontinuity here,
this jump discontinuity. So let's approach, let's
go from the positive, from values larger than three. So when x is equal to five,
g of five is a little bit more negative than negative three. G of four is between negative
two and negative three. G of 3.5 is getting a bit
closer to negative two. G of 3.1, it's getting even closer, closer to negative two. G of 3.01 is even closer to negative two. So it looks like this
limit right over here, well, I'm circling the wrong one, it looks like this limit exists. In fact, it looks like it
is approaching negative two. So this right over here
is equal to negative two. The limit of g of x as x approaches three from the right-hand side, and I'll just think about
it from the left-hand side. So I can start here. G of one, looks like it's a little bit greater than negative one. G of two, it's less than one. G of 2.5 is between one and two. G of 2.9 looks like it's a
little bit less than two. G of 2.99 is getting even closer to two. G of 2.99999 would be even closer to two so it looks like this
thing right over here is approaching two. So both of these limits,
the limit from the right and the limit from the left exist. The limit of g of x as x
approaches three exists. So these are the one-sided limit. This is the actual limit. Now in order for this
to exist, both the right and left-handed limits need to exist and they need to approach the same value. Well, this first statement, we saw that both of these exist
but they aren't approaching the same value. From the left, we are, or sorry, from the right, we are approaching, we are approaching negative two. And from the left, we are approaching two. So this limit does not exist. So I will not check that out
or I will not check that box. G is defined at x equals three. Well, when x equals
three, we see a solid dot right over there. And so it is indeed defined. It is indeed defined there. G is continuous at x equals three. Well, in order for g to be
continuous at x equals three, the limit must exist there. It must be defined there, and the value of the function there needs to be equal to
the value of the limit. Well, the function is defined there, but the limit doesn't exist there so it cannot be continuous. It cannot be continuous there. So I would cross that out. And I can't click, I wouldn't
click none of the above because I've already checked something, or I've actually checked
two things already.