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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition)>Unit 1

Lesson 7: Continuity at a point

# Worked example: Continuity at a point (graphical)

AP.CALC:
LIM‑2 (EU)
,
LIM‑2.A (LO)
,
LIM‑2.A.2 (EK)
Sal gives two examples where he analyzes the conditions for continuity at a point given a function's graph.

## Want to join the conversation?

• what is the specific difference between lim at x=3+ or 3- and 3 itself? and why?
(4 votes)
• When we write 3+ or 3‾ we meant one-sided limit. 3+ means x approaches 3 from the positive (right) direction and 3‾ means x->3 from the negative (left) direction.

When we just write 3 without the + or - we are talking about two-sided limit. Mean x approach 3 from both the left and right at the same time.

For a limit (two-sided) to exist, the limit from the left must equal to the limit from the right.

So for example,

Lim x->3+ = 2 and lim x->3- = 10, since they are not the same, then lim x->3 doesn't exist.
However, if lim x->3+ = 5 and lim x->3- = 5, since they are equal, then the lim x->3 exist and is 5.

Hope that helps.
(18 votes)
• I don't know if anyone has pointed this out yet but in the videos Sal says that for the graphs with asymptotes that no limit exists which is fair as from the left in the graphs we go through the limit is infinity which isn't a number and in the "classical" definition thus there is no limit. The problem is the questions ask the same thing in the quiz before this section but they want me to input the limit is infinity, in fact the answers available do not provide the "does not exist" option at all so Sal says one thing and the questions say another which is conflicting and confusing.
(5 votes)
• Perhaps you're right and the question and the tutorial are inconsistent.

Know this however: you are correct, infinity is not a number, and it cannot used as the value of a limit. [It's not a number in the sense that 2 is a number, let's say]
If a function approaches infinity (either positive or negative) as the x-value approaches the asymptote (from the left or right) then the function is unbounded at that point as much as it is undefined at that point. Indeed the "limit" does not exist. However the function can reach any arbitrarily large value for any suitably chosen x value (respective of left or right) close to the asymptote. Additionally note that the value of the function exactly at the asymptote is not defined.

In rational functions, you can have asymptotes or a "removable" discontinuity. Assuming we are dealing with asymptotes, approaching the x value of the asymptote from either the left or right will result in either a positive or negative infinity. Here the limit doesn't "equal" infinity, so much that it "does not exist". In the questions, they likely only want you to stipulate whether or not a function approaches positive infinity or negative infinity as x approaches the discontinuity. Feel free to substitute Lim f(x) "approaches" for "equals".

I hope these paragraphs help you.

Side-note: In future questions, I discourage you from using run-on sentences. The clarity of an answer can sometimes depend on the clarity of the question.
(7 votes)
• what does a solid dot and a hollow dot indicate??
(4 votes)
• A solid dot on a line indicates that the line goes through that point. A hollow dot on a line indicates that the line is not defined at that point.
(6 votes)
• At , how come g(x) is defined when x=3+ amd x=3- are not equal?
(4 votes)
• A function does not need to be continuous to be defined. When x = 3 there is a coordinate point that is defined that satisfies the situation, even though the limit does not exist.
(5 votes)
• What if a function is continuous before a point, but it's not defined immediately after that point ?
(3 votes)
• That's not uncommon. Take the function f(x)=x² on the interval [-1, 1]. f is continuous on that entire interval, including at the endpoints, but not defined past them.

You can also take this function and change the output at the points -1 and 1 only, so that the function is continuous on (-1, 1), discontinuous but still defined at -1 and 1, and undefined elsewhere.
(3 votes)
• How do I tell whether a function is defined at a point?
(2 votes)
• If you place the value of that point into the function, and the function solves into some real number, then the function is defined at that point.
(5 votes)
• Couldn't the second graph be continuous AT x=3? Since the point at y=2 isn't defined, wouldn't it be possible to say that when x=3, there's only one defined point and therefore the graph is continuous at that specific point?

edit: the reason I'm confused about this is because the question doesn't ask if the graph is continuous at the limit as x approaches 3 of g(x), but whether or not the graph is continuous at x=3. I completely understand that the limit is discontinuous but i don't understand how x=3 is.
(3 votes)
• 𝑔(2.9999...) = 2

But 2.9999... = 3, and 𝑔(3) = −2

So, even though we have chosen to define 𝑔(3) as equal to −2 there is a discontinuity at 𝑥 = 3.
(3 votes)
• The limit exists at g(x) when approaching x from 3- side and the limit is suppose to be 2, but how can it be the limit if at 2 the point is open.
(2 votes)
• If a certain point is not defined, but the values approaching it are, the limit still exists because when you find the limit of a certain value, you are finding the value the function approaches as x approaches that value. The function does not have to be defined at that value as long as the value of the function as x approaches that point exists (and, if you are finding a two-sided limit, the values as you approach x from both sides are equal).
(3 votes)
• For a limit to be continuous, shouldn't the right side and the left side should be defined and also should exist without any breaks. Right?
(2 votes)
• A limit is not continuous or discontinuous. For a limit to exist, it's necessary that the limits from left and right are both defined and equal each other.

For a function to be continuous at a point, its limit must exist at that point, and the function must equal the limit.

Hope that answers your question. Comment if it doesn't.
(3 votes)
• If a function begins at x=1 and is continuous ahead, can we call it continuous at 1?
(2 votes)
• Actually, we can't.

There are three conditions that must be met for a point on a graph to be continuous (I'll provide counterexamples for each condition).

1. The function must be defined at that point.
-This is straightforward. If the function is not defined at a point, it does not even exist there.
COUNTEREXAMPLE: Checking for point continuity at x=0 for a function only valid for x>5.
2. The limit of the function approaching the point in question must exist.
-The graph must connect. If the right and left-handed limits are different (or don't exist), the graph has two separate branches.
COUNTEREXAMPLE: Evaluating x=0 for f(x)={x when x>0; 5 when x<= 0}. The function is defined, but the limit does not exist.
3. The value of the defined point and the limit of the function approaching that point must have the same value.
COUNTEREXAMPLE: Evaluating at x=0 for f(x)={5 when -infinity<x<infinity; 0 when x=0}

In your case, the function you mentioned violates the 2nd and 3rd condition, as the left-handed limit cannot exist if the function begins at x=1 (If condition 2 is not met, 3 is also void).
(3 votes)

## 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.