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## Calculus, all content (2017 edition)

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

Lesson 16: 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?" •  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.)
• 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? • 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
• at , 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. • 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.
• where can I find a lesson or a course about ln, e and all of that algebric stuff? • 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? • 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.
• How can we tell by looking at the function that it is continuous at the given value? • 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?   