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## Calculus, all content (2017 edition)

# Discontinuity points challenge example

Based on out definition of continuity, we can see the relationship between points of discontinuity and two-sided limits. Created by Sal Khan.

## Want to join the conversation?

- I am just not understanding the concept of limits. why do we use them? what do limits mean?(13 votes)
- Limits have many uses. You could have an equation demonstrating the power a car's engine is able to put out in comparison to friction. The limit of this function,could easily be the maximum velocity the car could accomplish. You could have an equation relating the number of work hours logged by an employee in a day as compared to productivity. Because such a function would have a limited domain, it may be true that none of the local minimums and maximums you would find through derivative tests would be the overall min or max of that equation in that domain, thus you would need to compare them to the limit in order to maximize your efforts. Basically, limits have many uses, although they are most commonly used to test the maximum capability or productivity of people and machines, so as to get the best use out of them. Each one means something a little different depending on the context of the equation, but with a little critical thinking you can come up with that answer. I hope I've helped clarify this topic for you a little bit.(49 votes)

- I am just wondering whether the statement limit does not exist implies discontinuity?(7 votes)
- By definition a limit exists if the limit from the right and the limit from the left approach the same value. Therefore, if a limit does not exist, either;

the left-handed limit and the right-handed limit approach two different numbers - which would be a jump discontinuity or

they approach infinity or negative infinity, implying an asymptote, which also causes a discontinuity.(15 votes)

- If lim f(x) (x → k) does not exist, do we have to check the point of discontinuity k?(4 votes)
- Nope. In this problem we're looking for a value of k for which the limit exists, f(k) exists, but their two values are different (so that the function is not continuous at x = k). Therefore if lim f(x) (x → k) does not exist in the first place, we know we are not looking at that value of k here.(3 votes)

- Why is k equal to 8? Does it not represent the y-axis values? I would have thought that it would be equal to 7.(2 votes)
- This problem asks a question about the limit as x approaches k -- in other words, what number is x approaching when these conditions are satisfied. The answer has to be in terms of x because it's a question about what number x is approaching.(6 votes)

- how to determine the value of x for which the the function is discontinuous and the only given is the function f(x) and there is no graph?(1 vote)
- There are three primary sources of discontinuity:

1. A point where a piecewise function changes and there is a sudden jump in value. For example: f(x) = 2x where x < 2, and 400x³ ≥ 2

is discontinuous as x = 2.

2. A point where the function is not defined or fails to exist (such as division by zero).

3. A point where the function switches from having a real value to a complex value.(5 votes)

- Okay, here's an odd case: What about 1/x? 1/x is not defined at 0, but the limit of 1/x as x -> 0 is ALSO not defined. Or, rather, it doesn't exist. Does this mean that 1/x qualifies as continuous, or are "function is not defined" and "limit does not exist" considered different things? My intuition says "1/x is not continuous" simply because, well, just look at it.

Is there ANY function that's undefined at a given point that's still considered continuous, or must a function be defined?(2 votes)- The definition of continuity at a point requires that the function be defined at said point. Hence, if the function is not defined at a point, neither is it continuous there.(2 votes)

- So, if both sided limits exist, then limit exist, but it is just not continuous at that point k..?(1 vote)
- Almost. The limit exists if both of the one-sided limits exist AND are exactly equal. The function itself does not have to equal what the limits equal or even be defined.

Remember that the limit of a function at a particular point IS NOT necessarily the same value as the function itself. In fact, the limit can exist where the function is undefined OR the function can exist but the limit fail to exist at that point.(3 votes)

- Do limits exists in our physical world?(2 votes)
- I dont' understand why you can plug in a number when it is not on the constraint. Here, look at this. X<4, why can you plug 4 into the limit and then seemingly solve when 4 isn't even part of the domain restrictions to see where this limit is approaching?

I am sorry I don't have a specific limit, but for example, lim of x-4 as x goes to 4 when x<4.(2 votes) - At x=8 the value of the function is 7 but limit of that function is 1. This is fine a/c to limit but what exactly it means that in practical application or where can we see such type of function or why is that function like that and if the function is like that, than why we have to go too near to that point and define something(i.e., limit) even though we get huge difference in the answer. Plz explain as simple as possible!(2 votes)

## 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.