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Limits of combined functions

AP.CALC:
LIM‑1 (EU)
,
LIM‑1.D (LO)
,
LIM‑1.D.1 (EK)
,
LIM‑1.D.2 (EK)
Sal solves a few examples where the graphs of two functions are given and we're asked to find the limit of an expression that combines the two functions. For example, the limit at 0 of the product of the functions.

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  • aqualine ultimate style avatar for user Shinji 7800
    Hello,

    I may have a strange question but if I understand well "Limits" in maths ...

    x->n means x approach "n" but never reach it. So in the last exemple how can we say the limit is "4 over 0" and say the limit does not exist ?

    What am I missing ?

    best regards

    (Please forgive my approximate English ... I'm French)
    (41 votes)
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    • duskpin sapling style avatar for user Vu
      There are several ways to look at this.

      First, the limit law for quotient says that the lim_(x->a) F(x)/G(x) = lim_(x->a) F(x) / lim_(x->a) G(x) if lim_(x->a) G(x)≠0. In the example, lim_(x->a) G(x) = 0, therefore it doesn't exist.

      Second, dividing by 0 is undefined. So it doesn't exist.

      Third, x->0 means x is infinitely close to 0 but never 0. From this we can see that h(x) -> 4 (i.e. 3.9999999...) and g(x) ->0 (e.g. 0.0000001...). Now, when you divide a number by a very small number that is infinitely close to 0, you will get an infinitely large number. In other word, h(x)->4 / g(x)->0 will give us ∞. This will give us an infinite limit. An infinite limit does not exist because ∞ is not a number.
      (54 votes)
  • leaf green style avatar for user Alex Holden
    What is the difference between undefined, unbound, and does not exist?
    (14 votes)
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    • old spice man green style avatar for user Jonathan Ziesmer
      Consider the example of f(x)=0.5x+2 with the interval[2,∞):
      https://www.khanacademy.org/computer-programming/fx-05x-2-2/5706788113022976

      The limit as x -> ∞ of f(x) is ∞, right? But technically, infinity is unquantifiable. We know its a huge number, but by definition we can never find out exactly what its value is. So it is undefined.

      Now take the limit as x -> 1 of f(x). You can't because I've set the interval to be [2,∞). It simply does not exist in our function with this domain.

      Now consider the interval itself: [2,∞)
      Because there is a bracket ([) next to the 2, 2 is the left-most number our function can reach. f(x) is bounded in the negative direction by 2.
      Because the interval after the comma is followed by a parentheses ), it means f(x) will never actually reach the x coordinate of ∞. That means our function is unbounded in the positive direction.

      I hope this helps!
      (34 votes)
  • female robot ada style avatar for user Michaela Brock
    I'm sure this is probably really obvious, but is there a practical reason when you would add, subtract, multiply, or divide limits? Where might I see this applied?
    (18 votes)
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    • duskpin ultimate style avatar for user programmer
      i'm late by 3 years, but still I can answer with this example:
      In video games things have hitboxes and that is what is the actual thing, not the image showing up. Let's say we want the area of where to shoot the player, or rather the hitbox. For this with the help of limit relating to the hitbox/player's width + height will solve it.
      (9 votes)
  • leaf orange style avatar for user Prasid Sarkar
    In the third example how we can plot h(x) over g(x) graphically?
    (6 votes)
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    • aqualine ultimate style avatar for user That One Kid
      To my knowledge at the moment, in order to plot h(x)/g(x) you would first need the formulas for both. The example in this video does not have the formula. In other words, you cannot graph h(x)/g(x) with the information given.
      EDIT: (Thanks to Eric and stolenunder)
      I have just read that you actually could do it, albeit only for individual points, not for entire functions. Better explanations in the comments below.
      (4 votes)
  • male robot donald style avatar for user Sameer Raj Singh
    At , the limit of f(x) does not exist as x approaches to 0 since it is open at x=0 i.e., f(0) does not exist still you have considered the limit exist and taken its value as -1. Why?
    (3 votes)
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    • duskpin ultimate style avatar for user ☕
      This is because, it is apparent that as x approaches 0, the function would output -1 IF it were continuous. This is an assumption that has to be made when calculating limits of any function. Though remember that when calculating the limit at a jump discontinuity you can only find the limit from above or below. Cheers!
      (4 votes)
  • blobby green style avatar for user machetedeluz
    Hello guys! Shouldn't the final result be = ∞?

    I mean, if the numerator approaches a constant and the denominator approaches zero, the fraction should approach infinity, right?

    For instance:
    http://www.wolframalpha.com/input/?i=lim+(4%2Bx)%2F%7Cx%2F4%7C+x-%3E0
    (Copy the full URL in order to open the link properly.)

    Any help is welcomed :)
    (3 votes)
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  • blobby green style avatar for user P
    It was my understanding that the limits of functions at sharp or drastically changing points do not exist? is that right or wrong?
    (2 votes)
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  • duskpin tree style avatar for user S D
    When applying limit rules for f(x) and h(x) for example, do the functions have to both be continuous?
    (2 votes)
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  • duskpin tree style avatar for user Radek
    whats the difference between combined and composite functions?
    (2 votes)
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    • leafers tree style avatar for user coby
      Combined functions are functions merged together using arithematic +×÷-

      Composite is where you insert the reflection of 1 function into the other.
      f(g(x))=?
      How to look at this:
      What is g(x)? g(x) is the y value when you insert an x into the equation [g(x) = (blah)]
      You then take that y value and insert it as the x value for the function f(x)
      (4 votes)
  • old spice man green style avatar for user Dania  Zaheer
    Define f(0), so that the function, f(x)=((1-x)^n )-1/ x , n epsilon N, is continuous at X=0
    I'm stuck on this question, because I'm still a little confused. the concept isn't very clear to me yet
    (2 votes)
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    • leaf green style avatar for user kubleeka
      The function, as given, is not defined at x=0, since the function involves dividing by x. You're being asked to extend the domain of the function by one point, by deciding what f(0) will be, in such a way that your function is continuous at 0. To ensure that the new function is continuous at 0, you need to have f(0)=lim f(x) as x goes to 0.

      So really, what the problem is asking is for you to find lim f(x) as x goes to 0.

      Also, by 'n epsilon N', I assume you mean n∈ℕ. That symbol is not an epsilon, 𝜀, it means 'the number n belongs to the set ℕ', or 'n is a natural number'.
      (3 votes)

Video transcript

- [Instructor] Let's find the limit of f of x times h of x as x approaches zero. All right, we have graphical depictions of the graphs y equals f of x and y equals h of x. And we know, from our limit properties, that this is going to be the same thing as the limit as x approaches zero of f of x times, times the limit as x approaches zero of h of x. And let's think about what each of these are. So let's first think about f of x right over here. So on f of x, as x approaches zero, notice the function itself isn't defined there. But we see when we approach from the left, we are approaching the, the function seems to be approaching the value of negative one right over here. And as we approach from the right, the function seems to be approaching the value of negative one. So the limit here, this limit here is negative one. As we approach from the left, we're approaching negative one. As we approach from the right, the value of the function seems to be approaching negative one. Now what about h of x? Well, h of x we have down here. As x approaches zero, as x approaches zero, the function is defined at x equals zero. It looks like it is equal to one. And the limit is also equal to one. We can see that, as we approach it from the left, we are approaching one. As we approach from the right, we're approaching one. As we approach x equals zero from the left, we approach, the function approaches one. As we approach x equals zero from the right, the function itself is approaching one. And it makes sense that the function is defined there, is defined at x equals zero, and the limit as x approaches zero is equal to the same as the, is equal to the value of the function at that point because this is a continuous function. So this is, this is one. And so negative one times one is going to be equal to, is equal to negative one. So that is equal to negative one. Let's do one more. All right, so these are both, looks like continuous functions. So we have the limit as x approaches zero of h of x over g of x. So once again, using our limit properties, this is going to be the same thing as the limit of h of x as x approaches zero over the limit of g of x as x approaches zero. Now what's the limit of h of x as x approaches zero? This is, let's see, as we approach zero from the left, as we approach x equals zero from the left, our function seems to be approaching four. And as we approach x equals zero from the right, our function seems to be approaching four. That's also what the value of the function is at x equals zero. That makes sense because this is a continuous function. So the limit as we approach x equals zero should be the same as the value of the function at x equals zero. So this top, this is going to be four. Now let's think about the limit of g of x as x equals, as x approaches zero. So from the left, it looks like, as x approaches zero, the value of the function is approaching zero. And as x approaches zero from the right, the value of the function is also approaching zero, which happens to also be, which also happens to be g of zero. G of zero is also zero. And that makes sense that the limit and the actual value of the function at that point is the same because it's continuous. So this also is zero, but now we're in a strange situation. We have to take four and divide it by zero. So this limit will not exist 'cause we can't take four and divide it by zero. So even though the limit of h of x is x equals, as x approaches zero exists and the limit of g of x as x approaches zero exists, we can't divide four by zero, so this whole entire limit does not exist, does not exist. And actually, if you were to plot h of x over g of x, if you were to plot that graph, you would see it even clearer that that limit does not exist. You would actually be able to see it graphically.