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## Differential Calculus

### Course: Differential Calculus > Unit 1

Lesson 5: Properties of limits- Limit properties
- Limits of combined functions
- Limits of combined functions: piecewise functions
- Limits of combined functions: sums and differences
- Limits of combined functions: products and quotients
- Theorem for limits of composite functions
- Theorem for limits of composite functions: when conditions aren't met
- Limits of composite functions: internal limit doesn't exist
- Limits of composite functions: external limit doesn't exist
- Limits of composite functions

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# Theorem for limits of composite functions

AP.CALC:

LIM‑1 (EU)

, LIM‑1.D (LO)

, LIM‑1.D.1 (EK)

, LIM‑1.D.2 (EK)

Suppose we are looking for the limit of the composite function f(g(x)) at x=a. This limit would be equal to the value of f(L), where L is the limit of g(x) at x=a, under two conditions. First, that the limit of g(x) at x=a exists (and if so, let's say it equals L). Second, that f is continuous at x=L. If one of these conditions isn't met, we can't assume the limit is f(L).

## Want to join the conversation?

- 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.(88 votes)
- 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.(104 votes)

- how is lim g( f(x) ) can be changed to g( lim f(x) )? Khan please help us(33 votes)
- I couldn't find in the "Limit Properties" video mention of the rule which allows to decompose

into`lim f(g(x))`

, is it mentioned somewhere?`f( lim g(x))`

(25 votes)

- At the starting how did the intuition of opening the bracket work??(24 votes)
- 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.(17 votes)

- Does Khanacademy have some videos on the proof of these properties ?(15 votes)
- 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)- 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)

- 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 :)(11 votes)- 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)

- What is the difference between "undefined" and "does not exist"?(7 votes)
- 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)

- At4:02Sal 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)
- 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)

- after we have defined the limit of h(x) why is sal not looking for the limit of g(x)(5 votes)
- I think we don't have to because by defining h(x) we already have the answer to g(x) and that's what we're looking for in this type of exercise.

Look at1:43(2 votes)

- At4:55, wouldn't h(1) be "all real numbers" since there are an infinite number of y-values when x is equal to 1?(4 votes)
- That's not quite how infinity works. There is still technically only one y value, the y value is just THE smallest y value there is. there is no largest or smallest value though so we have to say it is undefined.

Let me know if that didn't make sense.(4 votes)

## 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.