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Limits of combined functions: piecewise functions

This video demonstrates that even when individual limits of functions f(x) and g(x) don't exist, the limit of their sum or product might still exist. By analyzing left and right-hand limits, we can determine if the limit of the combined functions exists and find its value.

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  • leafers ultimate style avatar for user david.rau
    In this video, it seems that main idea interferes with the limit properties video previously, and the traditional view of limits learned through most calculus texts. The general idea is the limit of sum is the same as the sum of the limits... If we then take the sum of each limit, each limit is DNE so how can we combine them in the traditional sense?
    (39 votes)
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    • female robot grace style avatar for user Isaac Kriegerventh
      The idea about the existence of the limit of a function at any value "p" is that the one sided limits as x -> p are equal. If we make the graph of the combined functions showed in the video we will see that the one sided limits are equal in the first and third case but not in the second. There will be a discontinuity when the limit doesn't exist.
      (20 votes)
  • blobby green style avatar for user mohamad
    i can pass the test about this video but i will never understand the proof of this.can s1 explain?
    (10 votes)
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    • blobby green style avatar for user carlosfermin38
      By combining f(x) and g(x) you are basically creating a new function that does not necessarily follow the same principles as the original functions.

      Take for example:
      f(x) = 1/(x-1)
      g(x) = (x-1)^2
      h(x) = f(x)*g(x) = ((x-1)(x-1))/(x-1) = x-1

      If we wanted to calculate the limit of x->1 of f(x) we would get:
      1/1-1 = 1/0, which is undefined

      However, if we take this same limit from h(x) we would simply get 0, as the new function is x-1
      (4 votes)
  • blobby green style avatar for user s345732
    Can the limit ever exist if one function is continuous while the other isn't? I'm thinking if you had the same number for both the left approach and the right approach (from the continuous function) while having different numbers added (from the piece wise) you'd have different numbers for the left and right approach since you'd be adding a different number to the same number so you'd always come out with DNE for a couple of functions where one is continuous and one isn't?
    (6 votes)
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    • male robot donald style avatar for user Venkata
      That's a really good question! Had me thinking for a solid 1 hour (mostly because I couldn't find more than one example to disprove it lol!)

      There is a special case where if one function is continuous and the other isn't, the sum of limits will still exist. This only happens (as far as I know) with piecewise functions with point discontinuities. For example, if I have a continuous function f(x) and a discontinuous function g(x) which has a point discontinuity at x = 3, the limit as (f(x) + g(x)) tends to 3 will still exist, as the only difference between f(x) and g(x) is that in g(x), x=3 isn't in its domain.

      In general though, the limit will indeed not exist, as, just like you said, you're adding/subtracting different numbers from the discontinuous function to the same number in the continuous one, hence yielding different numbers and thus, a non-existent limit.
      (11 votes)
  • blobby green style avatar for user Adil Adil Rehimli
    i cant understand if the limits dont exist how can we sum them ?
    (8 votes)
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  • blobby green style avatar for user kmchen
    When dealing with combined functions, why do we need to check the limit as it approaches from the left side and right side?

    Whereas in the later on composite function video, we do not have to check the limit as it approaches from the left side and right side? It makes sense, however I am having difficulty putting it into words.
    (5 votes)
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    • purple pi pink style avatar for user Sebastian
      For a limit to exist, it has to a approach a certain value from both the right and left side, and when you have the limit of the sum or product of two functions, even if the limit of each function doesn't exist, the limit of the sum/product might still exist if it meets this condition.

      Also, in the later video, Sal is taking the limit from the right and left side, if from both sides it approached a different value the limit wouldn't exist.
      (7 votes)
  • primosaur ultimate style avatar for user Qihang Yao
    So, in short, it doesn't matter if the individual limits exist, if the left and right of lim(f(x) + g (x) ) equal, there is a limit?
    (7 votes)
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    • starky ultimate style avatar for user Ray2017
      The individual limit does exist. Take x -> -2 (f(x) + g(x)) for example. Think of (f(x) + g(x)) as a single function that can be represented by f(x) and g(x). If you combine them, you will realize both the limits approaching from the right and left are 4. So in general, view whatever inside the parenthesis as a single function THEN take the limit.
      (3 votes)
  • aqualine ultimate style avatar for user Julien Lee Heberling
    At minute , as x approaches -2 from the positive side, doesn't the y value approach 3? The video says that the y value approaches 2 when x approaches -2 from the positive side. Is this a mistake, or am I missing something? Many thanks,
    (1 vote)
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  • blobby green style avatar for user okodieq
    With respect to the first example, the question was straightforward, and the limit was undefined. Now, must we always prove the limit exists even if the question does not ask that?
    (1 vote)
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  • blobby green style avatar for user madmonsterhi
    why does dividing with 0 result in DNE but multiplying by 0 does not?
    (1 vote)
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  • blobby green style avatar for user Caleb Symmonds
    This video was so confusing just because Sal mixed up left and right so many times. If you're approaching FROM the left-hand side, you're going TOWARDS the right-hand side. From the left to the right. From the right to the left.

    Sal went from the right to the right and from the left to the left, which didn't make any sense. This video would be helpful with some edits, or even just doing the video over again.
    (1 vote)
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Video transcript

- [Instructor] We are asked to find these three different limits. I encourage you like always, pause this video and try to do it yourself before we do it together. So when you do this first one, you might just try to find the limit as x approaches negative two of f of x and then the limit as x approaches negative two of g of x and then add those two limits together. But you will quickly find a problem, 'cause when you find the limit as x approaches negative two of f of x, it looks as we are approaching negative two from the left, it looks like we're approaching one. As we approach x equals negative two from the right, it looks like we're approaching three. So it looks like the limit as x approaches negative two of f of x doesn't exist, and the same thing's true of g of x. If we approach from the left, it looks like we're approaching three. If we approach from the right, it looks like we're approaching one. But it turns out that this limit can still exist as long as the limit as x approaches negative two from the left of the sum, f of x plus g of x, exists and is equal to the limit as x approaches negative two from the right of the sum, f of x plus g of x. So what are these things? Well, as we approach negative two from the left, f of x is approaching, looks like one, and g of x is approaching three. So it looks like we're approaching one and three. So it looks like this is approaching. The sum is going to approach four. And if we're coming from the right, f of x looks like it's approaching three and g of x looks like it is approaching one. Once again, this is equal to four. And since the left and right handed limits are approaching the same thing, we would say that this limit exists and it is equal to four. Now let's do this next example as x approaches one. Well, we'll do the exact same exercise. And once again, if you look at the individual limits for f of x from the left and the right as we approach one, this limit doesn't exist. But the limit as x approaches one of the sum might exist, so let's try that out. So the limit as x approaches one from the left hand side of f of x plus g of x, what is that going to be equal to? So f of x, as we approach one from the left, looks like this is approaching two. I'm just doing this for shorthand. And g of x, as we approach one from the left, it looks like it is approaching zero. So this will be approaching two plus zero, which is two. And then the limit, as x approaches one from the right hand side of f of x plus g of x is going to be equal to. Well, for f of x as we're approaching one from the right hand side, looks like it's approaching negative one. And for g of x as we're approaching one from the the right hand side, looks like we're approaching zero again. Here it looks like we're approaching negative one. So the left and right hand limits aren't approaching the same value, so this one does not exist. And then last but not least, x approaches one of f of x times g of x. So we'll do the same drill. Limit as x approaches one from the left hand side of f of x times g of x. Well, here, and we can even use the values here. We see it was approaching one from the left. We are approaching two, so this is two. And when we're approaching one from the left here, we're approaching zero. We're gonna be approaching two times zero, which is zero. And then we approach from the right. X approaches one from the right of f of x times g of x. Well, we already saw when we're approaching one from the right of f of x, we're approaching negative one. But g of x, approaching one from the right, is still approaching zero, so this is going to be zero again, so this limit exists. We get the same limit when we approach from the left and the right. It is equal to zero. So these are pretty interesting examples, because sometimes when you think that the component limits don't exist that that means that the sum or the product might not exist, but this shows at least two examples where that is not the case.