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Theorem for limits of composite functions

This video focuses on finding the limit of composite functions, specifically the limit as 'x' approaches 'a' of f(g(x)). It explains that this limit equals f(limit as 'x' approaches 'a' of g(x)) if two conditions are met: the limit of g(x) exists, and f(x) is continuous at that limit.

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  • blobby green style avatar for user Stas kaufman
    Which "Limit property" allowing us to do this bracket "jump", I looked at the first video in this section and didn't figure it out.
    (91 votes)
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    • scuttlebug green style avatar for user Lars
      There's a confusing omission here in the video series, because (1) the "composition law" is assumed here but was never presented ("We can leverage our limit properties," the video says, but that property was left out); and (2) the extra condition for the composition law, namely that g(x) must be continuous at L, isn't mentioned. Nor do any of these examples bring up that extra condition. Instead, students are left to stumble over this condition by answering some of the following exercises wrong and then trying to figure out why they were wrong.

      So, two recommendations: (a) Add the composition law to the "Limit properties" video, including the requirement for g(x) to be continuous at L; and (b) add to this video an example where the only thing stopping us from being able to evaluate the limit is that h(x) is not continuous at L. I.e. lim (x->L) h(x) must exist, but be different from h(L). Examples 2 and 3 in this video already fail to meet the extra condition, but they also fail to meet earlier conditions and so we don't get to see the continuity condition in action.
      (130 votes)
  • primosaur seedling style avatar for user FredMun
    how is lim g( f(x) ) can be changed to g( lim f(x) )? Khan please help us
    (35 votes)
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  • blobby green style avatar for user avy1234
    At the starting how did the intuition of opening the bracket work??
    (25 votes)
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    • duskpin ultimate style avatar for user Igor
      What Sal didn't mention in this video (but is explained later in the following exercise):
      If functions h(x) and g(x) both have limits and and g(x) is continuous at x=L, [in other words ( lim(x→c)h(x) = L and lim(x→c)g(x) = g(L)] then
      lim(x→c) g(h(x)) = g(lim(x→c)h(x)) = g(L)
      So, if I'm not mistaken, since "outer" function g(x) should be continuous (in order for this property to hold) at the given limit then lim(x→c)g(L) = g(L), since function is defined at the limit, the limit is equal to function's output.
      (20 votes)
  • starky ultimate style avatar for user Andres Romero
    Does Khanacademy have some videos on the proof of these properties ?
    (21 votes)
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    • leaf green style avatar for user cossine
      Proofs involving limits are outside the scope of high school math. If you are interested you should learn the basics of discrete maths, logic, set theory and proof techniques.

      From there you should study a real analysis textbook (this is just calculus), where you find proof of the theorem mentioned in this video. You will also find proof of the other limit properties mentioned.
      (2 votes)
  • blobby green style avatar for user Ignacio.Cryptmaw
    lim { x→1 } f ( g(x) ) = f ( lim { x→1 } g(x) ),
    so, does that mean that this whole limit will exist as long as lim {x→1} g(x) is defined and f(that value of g) is defined, even if lim {x→g} f(x) does not exist?

    For example, g(x) approaches 3 when x approaches 1, and f(3) = 10 but the function f(x) is discontinuous at f(3) such that the one side limits are different and hence its limit is undefined, will lim {x→1} f(g(x)) return the value 10?
    (11 votes)
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    • leaf grey style avatar for user Qeeko
      No, and here is a counterexample. Define g(x) = 3 - |x-1|. Observe that g(x) → 3 as x → 1. Define the function ƒ as follows: ƒ(x) = 0 if x < 3 and ƒ(x) = 10 if x ≥ 3. For any x ≠ 1, one has g(x) < 3, and so ƒ[g(x)] = 0 for such x. It follows that

      lim (x → 1) ƒ[g(x)] = lim (x → 1) 0 = 0 ≠ 10 = ƒ(3).
      (10 votes)
  • aqualine ultimate style avatar for user Shinji 7800
    Hello, I think an example is missing.
    Imagine lim (x -> -2) of f(g(x))
    lim (x -> -2) g(x) is 0

    f(0) = -1 but is not continuous such as lim ( x -> 0-) f(x) = 1 and lim ( x -> 0+) f(x) = -1

    but f(0) is defined and equal -1 (in fact is just a discontinuity

    what is the limit ? it is defined ?

    thanks :)
    (12 votes)
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    • starky ultimate style avatar for user yury
      I thought just like you, but in one of the mastery challenges the answer like this was stated erroneous. Explanation from "hints" section says, that both limits should exist and f(x) is continuous at x. I'm a bit confused.
      (3 votes)
  • female robot grace style avatar for user bad
    What is the difference between "undefined" and "does not exist"?
    (7 votes)
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    • leaf green style avatar for user jude4A
      It's more of a semantic difference than a mathematical difference. I've only seen "does not exist" in the context of limits and their existence. I've seen undefined used in expressions (such as 0^0, infinity-infinity) and slopes of a line.
      (7 votes)
  • leaf red style avatar for user Jack McClelland
    At Sal evaluates the function at the point -3 is approaching the value of 1, but there is a discontinuity, a point that looks like it's at -3, 4. Why is the limit not 4?
    (7 votes)
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    • duskpin ultimate style avatar for user Igor
      The limit, by definition, is a point that function is approaching as the x-values approach a specific value. In this case function's output at the x = -3 and lim( x→ -3) of f(x) will be different values. Just like in other cases when we have point discontinuity.
      (2 votes)
  • blobby green style avatar for user agam022005
    Is there a reputable source that also shows this proof? I need that for citing for my project. if someone can help it would be great. Thanks!
    (4 votes)
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  • blobby green style avatar for user F Lite
    after we have defined the limit of h(x) why is sal not looking for the limit of g(x)
    (5 votes)
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Video transcript

- [Tutor] In this video, we're going to try to understand limits of composite functions, or at least a way of thinking about limits of composite functions and in particular, we're gonna think about the case where we're trying to find the limit as x approaches a, of f of g of x and we're going to see under certain circumstances, this is going to be equal to f of the limit, the limit as x approaches a of g of x and what are those circumstances you are asking? Well, this is going to be true if and only if two things are true, first of all, this limit needs to exist. So the limit as x approaches a of g of x needs to exist, so that needs to exist and then on top of that, the function f needs to be continuous at this point and f continuous at L. So let's look at some examples and see if we can apply this idea or see if we can't apply it. So here I have two functions, that are graphically represented right over here, let me make sure I have enough space for them and what we see on the left-hand side is our function f and what we see on the right-hand side is our function g. So first let's figure out what is the limit as x approaches negative three of f of g of x. Pause this video and see, first of all, does this theorem apply? And if it does apply, what is this limit? So the first thing we need to see is does this theorem apply? So first of all, if we were to find the limit as x approaches negative three of g of x, what is that? Well, when we're approaching negative three from the right, it looks like our function is actually at three and it looks like when we're approaching negative three from the left, it looks like our function is at three. So it looks like this limit is three, even though the value g of negative three is negative two, but it's a point discontinuity. As we approach it from either side, the value of the function is at three. So this thing is going to be three, so it exists, so we meet that first condition and then the second question is is our function f continuous at this limit, continuous at three? So when x equals three, yeah, it looks like at that point, our function is definitely continuous and so we could say that this limit is going to be the same thing as this equals f of the limit as x approaches negative three of g of x, close the parentheses and we know that this is equal to three and we know that f of three is going to be equal to negative one. So this met the conditions for this theorem and we were able to use the theorem to actually solve this limit.