Main content

### Course: Differential Calculus (2017 edition) > Unit 3

Lesson 5: Limits of piecewise functions# Worked example: point where a function isn't continuous

Sal finds the limit of a piecewise function at the point between two different cases of the function. In this case, the two one-sided limits aren't equal, so the limit doesn't exist..

## Want to join the conversation?

- in real life application, what does it mean that a "limit does not exist?"(19 votes)
- It means that the function does not approach some particular value. Take sin(x) for example. It is defined for any x, but the limit of sin(x) as x goes to infinity does not exist, because it doesn't get closer to any value; it just keeps cycling between 1 and -1. Or take g(x) = (1/x)/(1/x). It is not defined at 0, but the limit as x approaches 0 (or any other value for that matter) is 1.

Note that whether the limit exists or not does not only depend on the function, but it also depends on the input. What does f(x) get close to as x gets close to _?

As for real life applications: Suppose you are measuring brightness as a function of time (say, in minutes after midnight January 1st 2000). This would be something like sin(t), and the earlier example applies. The sunrise/sunset cycle would mean that you could ask about the limit as you approach 8 in the morning tomorrow, but it would not make sense to ask about the limit as t ticks by, day after day. You'd just get infinitely alternating brighter/darker values. You could also have a discontinuity. Suppose you are measuring the brightness inside rather than outside. If you turn on the lights at some particular point in time, then it does not make sense to talk about the brightness in the room approaching that value (as t gets closer to when you flipped the switch). The function is defined at that point, but the graph looks very different on either side. (The limits as you get closer from the left or the right are different.)(38 votes)

- How do you come up with the limit of f of x as x is approaching c if you have three equations instead of two like in this video?(4 votes)
- Hi Karen,

These are piecewise defined functions and altogether, they create one function that is defined by different rules depending on where you are in the domain. If 'c' lives in one of the intervals defined in the function, then you will evaluate the limit of the piecewise function using the rule defined for that domain. If 'c' lives where two of the rules come together, then you'll use one rule for the left hand limit and the other for the right hand limit.

Piecewise functions can be created with any amount of rules, evaluating limits will always depend on wherever 'c' is defined(6 votes)

- at2:48, the clause is evaluated at 2.However, up there it says: for x>2.

So shouldnt it be evaluated with a value greater than 2 in this case?

in the end it doesnt say x is greater than or equal to 2, it says x is greater than 2.(4 votes)- It's evaluated as a limit as x approaches 2 from the right. As long as a function is defined infinitely close to 2, f(2) doesn't have to exist to take that limit. As another example, the limit as x approaches 1 of [(x+1)(x-1)]/(x-1) is just (1+1) or 2, even though f(1) doesn't exist. The closer you get to x=1, the closer the function gets to two.(3 votes)

- where can I find a lesson or a course about ln, e and all of that algebric stuff?(3 votes)
- I have some questions:

a. If the function f has a limit when x→a and function g doesn't have a limit. What can we say about f-g? b. If function g doesn't have a limit, but the function f/g has a limit when x→a, what can we say about the function f?(3 votes)- As Sal had explained in the properties of limits and the consequent videos:

a. When the limit of g does not exist, then even though limit of f exists; limit of f-g doesn't exist.

b. If function g does not have a limit at x=a, and function f/g has a limit at x=a, then the function f will be a factor of (x-a) or factor of function g, thus using rationalization, f/g will prove to have a limit. F can either have a limit as in eg: F=2/(x-a) and g= 1/(x-a) or have a limit as in f= x-a and g = 1/x-a

Kindly correct me if I am wrong.(4 votes)

- How can we tell by looking at the function that it is continuous at the given value?(3 votes)
- Intuitively, a function is continuous if you can draw it without picking up your pencil, it's a single connected line. If you have to pick up your pencil to accommodate a hole or a jump, then the function is discontinuous.(3 votes)

- Is it correct to say that "the limit is undefined" instead of "the limit doesn't exist"?(1 vote)
- So is the function continuous at x = 2?(2 votes)
- No. Because the limit does not exist, then it is not continuous there.(2 votes)

- Can you help me out with a concept similar to this. I need to find the values of parameters that would make a piecewise defined function continuous everywhere. However, the limit as x->1 does not exist as the values from each side are different, and when i substitute x=1 directly into the function, it is a different value as well. Therefore I am confused as to how to go about finding 2 values that would make the function continuous. Thanks for you help if you get to read this.(2 votes)
- honestly, this cleared up so much for me. thanks. ( better then my teacher)(2 votes)

## Video transcript

- [Voiceover] So we've
got this function f of x that is a piecewise continuous. It's defined over several intervals here for x being, or for zero less than x, and being less than or equal to two. F of x is natural log of x. For any x's larger than two, well then, f of x is going to be x squared times the natural log of x. And what we wanna do is we wanna find the limit of f of x as x approaches two. Now, what's interesting
about the value two is that that's essentially
the boundary between these two intervals. If we wanted to evaluate it at two, we would fall into this first interval. F of two, well, two is
less than or equal to two, and it's greater than zero. So f of two would be
pretty straightforward. That would just be natural log of two. But that's not necessarily
what the limit is going to be. To figure out what the
limit is going to be, we should think about, well, what's the limit as we
approach from the left, what's the limit as we
approach from the right, and do those exist, and if they do exist, are they the same thing, and if they are the same thing, well then, we have a well-defined limit. So let's do that. Let's first think about the limit. The limit of f of x as we approach two from the left, from values lower than two. Well, this is gonna be the case where we're gonna be
operating in this interval right over here. We're operating from values less than two, and we're going to be approaching two from the left. And so we'll fall under this clause, and so, since this clause or case is continuous over the interval in which we're operating, and for sure between, or for
all values greater than zero and less than or equal to two, this limit is going to be equal to just this clause evaluated at two, because it's continuous over the interval. So this is just going to
be the natural log of two. All right, so now let's
think about the limit from the right hand side, from values greater than two. So the limit of f of x as x approaches two from
the right hand side. Well, even though two
falls into this clause, as soon as we go anything
greater than two, we fall in this clause. So we're gonna be approaching two essentially using this case. And once again, this case here is continuous for all x values not only greater than two, actually, you know, greater
than or equal to two. And so, for this one over here, we can make the same
argument that this limit is going to be this
clause evaluated at two, because once again, if we just evaluated the function at two, it
falls under this clause, but if we're approaching from the right, if we're approaching from the right, those are x values greater than two, so this clause is what's at play. So we'll evaluate this clause at two. So, because it is continuous. So this is going to be two squared times the natural log of two. And so this is equal to four times the natural log of two. Four times the natural log of two. So, the right hand limit does exist. The left hand limit does exist. But the thing that might jump out at you is that these are two different values. We approach a different
value from the left as we do from the right. If you were to graph this, you would see a jump in the actual graph. You would see a discontinuity
occurring there. And so for this one in particular, you have that jump discontinuity, this limit would not exist because the left hand limit
and the right hand limit go to two different values. So this does not exist.