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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB > Unit 1

Lesson 5: Determining limits using algebraic properties of limits: limit properties- 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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# Limits of composite functions: internal limit doesn't exist

Finding the limit of g(h(x)) at x=-1 when the limit of h(x) at x=-1 doesn't exist. Does it mean that the composite limit doesn't exist? Not necessarily! See how we analyze it. Created by Sal Khan.

## Want to join the conversation?

- Shouldn't it be -3 from the left, so -3^- ?(36 votes)
- It also took me some time to understand. Basically, he is shortening the stuff.

When you take the limit of h(x) from the right-hand side (x=>1^+) you get -2.

When you take the limit of h(x) from the right-hand side (x=>1^-) you get -3.

After that when you do the same for g(h(x)) from the left and right-hand side for h(x=>-2), you will get 3 for both hand sides.

Same for the h(x)=>-3, both left and right-hand sides will be the same, with the answer of 3.

instead of doing the whole stuff he just did the desired and required stuff.(7 votes)

- It is confusing from3:05while taking the LH limit of g(h(x)) as x app-1^- and equating to LH limit of g(h(x)) as h(x) app -3 #should it be -3^- or -3^+. Please answer.(11 votes)
- It also took me a while to figure out:

When you're thinking about what h(x) is approaching, you are looking at the**y (vertical) axis**.

It'd be as if you tilt the whole image by 90º (to the right), or tilt your head to the left like an owl. And so, on the y axis:

"right" (+) is the upper side of the graph,

"left" (-) is the lower side of the graph.

So if you look at this graph, h(x) is approaching*both*-2 and -3 from the*right*/*upper*(+) side.

You don't have to look at*both*+ and - for each. You only have to figure out where it's approaching*from*.

Hope that helps!(20 votes)

- After an initial confusion, I think I have understood the procedure, but I would like confirmation that I have done it right, because I am not sure.

Towards the end, Sal calculates the limit of h(x) as x approaches -1 from the left. That is easy: -3.

But then the complicated thing comes, because to find the corresponding limit of g(x), looks for the limit from the right (!). What I think is the following:

We have -3 we got from the previous step. But if we look at the graph, the values of y are diminishing as they approach that point. That is, -3 is approximated from above.

Now, when we plug this value in the x-axis of g(x), we need to do it in the same way, that is, from above, i.e., from the values greater than -3, which means that we are approximating from the right.

Can anybody confirm that I have understood it right? Thanks.(9 votes)- Correct. I find it healthier to think of from the right as "values above x " i.e. x⁺ and from the left as "values below x " i.e. x⁻.(1 vote)

- Stuck at h(x) -> -3+, checked the comments here and on YT, people say its + and not - because we're looking at the Y values. I thinks that's right but how do we know which value to look at? In previous videos I was under the impression that we considered the X axis when determining right hand or left hand?(7 votes)
- I think considering the Y-axis is just for when we are doing composite functions because in a composite function, we are taking the
**outputs/y-values**of the**internal function**and then using them as**inputs/x-values**of the**external function**.

For other operations such as addition, subtraction, multiplication and division, the y-values**remained**as y-values, they never became x-values for another function which is why it was a lot more straight forward.

Here are steps that I found from someone else (I can't seem to find their comment anymore) that really helped me understand what was going on:

1) Take the limit of the function h(x) as x -> -1 from the right

2) The**OUTPUT**of that function will be -1.9, -1.99, -1.999...

3) That sequence of numbers will serve as the**INPUT**for g(x): g(-1.9), g(-1.99), g(-1.999)...

4) Following that same series of inputs in g(x) as x -> -2 means you get closer to -2 from the**RIGHT/POSITIVE**side of the graph

5) Do the same thing for the left side and. we see that x seems to be approaching -3 from both sides

Hope this helps and If I made a mistake anywhere please be sure to correct me!(4 votes)

- This is confusing. Inputting something that does not exist into a function should yield something that does not exist. How come it actually exists?(3 votes)
- The meaning of a limit that DNE just means that the one-sided limits are different i.e. value of the function as x approaches a from below is different from the value of the function as x approach a from values above a. But this is only true for the inner function h(x).

We know that the lim x→-1 g(h(x)) exists and is true so long if lim x→-1⁺ g(h(x)) = lim x→-1⁻ g(h(x)). We just need to prove that the one-sided limits for the composite function are the same for the limit of the composite function to exist.

The composite function is taking the output of the inner function as input.

As x→-1⁺ for h(x), the output h(x)→-2⁺. Plugging this h(x) output**as the x-input**into g(x), lim h(x)→-2⁺ g(h(x)) = 3

As x→-1⁻ for h(x), h(x)→-3⁺. Plugging this h(x) output**as the x-input**into g(x), lim h(x)→-3⁺ g(h(x)) = 3

To visualise this, I found it very helpful to visualise this on the graphs. Look at the video again, notice Sal uses the arrows in both h(x) and g(x).

As x→-1⁺, draw an arrow of h(x)→-2⁺. Now look at g(x) which takes h(x) as its x-input.

For g(x), as x (now x is h(x)) → -2⁺, g(x) = 3.

Now go back to h(x). As x→-1⁻, limit h(x) → -3⁻ and draw an arrow as h(x) → -3⁻.

Now go to g(x) which takes h(x) as its x-input, as x (now is equal to h(x)) → -3⁺, g(x) = 3

So we have proven that lim x→-1 g(h(x)) exists and is equivalent to: lim x→-1⁺ g(h(x)) = lim x→-1⁻ g(h(x)) because the one-sided limits are the same.

Avoid being overly fixated on the thought that the limit that DNE. DNE just means the one-sided limits are different. Just because the one-sided limits are different for one function does not mean that the one-sided limits of another function (which could take these different one-sided limits as input) are different.

Thus, even if a limit does not exist (one-sided limits are different) for one function, does not mean that the limit for another function who takes these different one-sided limits as input, does not exist.(3 votes)

- I'm in 6th grade, so this is taking a bit to comprehend.(3 votes)
`h(x)`

has no limit as it approaches the x-position of -1. From the left side, the function seems to be approaching y = -3, while from the left side, the function seems to be approaching y = -2. Since the "left and right limits" don't match,**the function doesn't have one single limit**.*However*, we are not focusing on the limit of`h(x)`

by itself, we are focusing on the composite function`f(h(x))`

. The limit of`h(x)`

from the left side is -3, so plugging that into our function`f(h(x))`

would result in`f(-3)`

from the left side, since`h(x)`

= 3 (as the function approaches -1 from the left).

The limit of`h(x)`

from the right side is -2, so that would be`f(-2)`

. In the graph`f(x)`

, the limit of f as it approaches -2 and the limit of`f(x)`

as it approaches -3 are both equal to 3, so the limit of the composite function`f(h(x))`

is equal to three.

I hope you understand! If I made anything unclear, please let me know. Also, nice job taking the initiative and learning Calculus in 6th grade (I am learning it early too)!(3 votes)

- What happens when you get two different answers for both the g(h(x))?(3 votes)
- Then the limit does not exist(2 votes)

- So why can you not just sub x = -1 into h(x), since the value is defined in the function? h(-1) is defined as -3, so finding lim x→-1 g(-3) is 3. Unless I'm missing something here?

I understand that if you are instead operating on g[lim x→-1 h(x)] then the limit doesn't exist. But it seems like an unnecessary step in this case. Sal even states in a previous video that:

lim x→-1 g(h(x)) = g[lim x→-1 h(x)], only if:

- lim x→-1 h(x) exists (it doesn't); and

- g(x) is continuous at L. Which when you work it out g(-3) = L is not continuous.

From what I've been researching, it seems that for lim x→-1 g(h(x)) to exist, the limit for h(x) must also exist. From study.com:*Find the limit of the inner function. If the limit of the inner function at a given value is undefined, then the limit of the composite function is undefined also.*

Does that means you cannot simply sub x into h(x), and you must instead use lim x→-1 h(x)? If this is the case, then it's a little confusing as I haven't seen that fact explained on Khan Academy.(2 votes)- lim 𝑥→−1 ℎ(𝑥) ≠ −3

In other words, 𝑥 → −1 ⇏ ℎ(𝑥) → −3

Thus, lim 𝑥→−1 𝑔(ℎ(𝑥)) ≠ lim ℎ(𝑥)→−3 𝑔(ℎ(𝑥)) = lim 𝑥→−3 𝑔(𝑥)

– – –

By the way...

"If the limit of the inner function at a given value is undefined, then the limit of the composite function is undefined also."

That is NOT necessarily true, which Sal basically showed in the video.

As another example, take the functions

ℎ(𝑥) = −1 for 𝑥<0, 1 for 𝑥≥0

𝑔(𝑥) = 𝑥²

Then 𝑔(ℎ(𝑥)) = 1

and lim 𝑥→0 𝑔(ℎ(𝑥)) = 1

even though lim 𝑥→0 ℎ(𝑥) is undefined.

If you could provide a link to that article I'd very much appreciate it.(3 votes)

- In the video I think there was a mistake. Where it is approaching -1 from the left h is approaching-3 the values are less not greater but the solution is right but the terminology is wrong. Am I right?(2 votes)
- The y part of the function is the same as h(x), is approaching -3 from values greater than -3 so i dont see a mistake(1 vote)

- Can anyone simplify this?(2 votes)
- Sure, Adi! Here's a simplified version of the video:

Sal is explaining how to find the limit of composite functions. He uses an example where he wants to find the limit of g(h(x)) as x approaches -1, where g and h are functions defined graphically. He points out that the limit of h(x) as x approaches -1 does not exist, as it approaches different values from the right and left. However, he shows that the limit of g(h(x)) still exists by considering the right-hand and left-hand limits separately. He determines that the right-hand limit and left-hand limit are both equal to 3, so the overall limit of g(h(x)) as x approaches -1 is also 3. Sal concludes that this is a unique example where the limit of the composite function exists even though the limit of the internal function (h(x)) does not exist.(1 vote)

## Video transcript

- [Instructor] All right, let's get a little bit more practice taking limits of composite functions. Here, we want to figure
out what is the limit as x approaches negative
one of g of h of x? The function g, we see
it defined graphically here on the left, and the function h, we see it defined graphically
here on the right. Pause this video and have a go at this. All right, now your first
temptation might be to say, all right, what is the limit
as x approaches negative one of h of x, and if that limit
exists, then input that into g. If you take the limit as
x approaches negative one of h of x, you see that
you have a different limit as you approach from the right than when you approach from the left. So your temptation might be
to give up at this point, but what we'll do in
this video is to realize that this composite limit actually exists even though the limit as
x approaches negative one of h of x does not exist. How do we figure this out? Well, what we could do
is take right-handed and left-handed limits. Let's first figure out what is the limit as x approaches negative
one from the right hand side of g of h of x? Well, to think about that,
what is the limit of h as x approaches negative one
from the right hand side? As we approach negative one
from the right hand side, it looks like h is
approaching negative two. Another way to think about
it is this is going to be equal to the limit as h of
x approaches negative two, and what direction is it
approaching negative two from? Well, it's approaching
negative two from values larger than negative two. H of x is decreasing down to negative two as x approaches negative
one from the right. So it's approaching from
values larger than negative two of g of h of x. G of h of x. I'm color coding it to be
able to keep track of things. This is analogous to
saying what is the limit, if you think about it as
x approaches negative two from the positive direction of g? Here, h is just the input into g. So the input into g is
approaching negative two from above, from the right I should say, from values larger than negative two, and we can see that g
is approaching three. So this right over here is
going to be equal to three. Now, let's take the limit
as x approaches negative one from the left of g of h of x. What we could do is first think
about what is h approaching as x approaches negative
one from the left? As x approaches negative
one from the left, it looks like h is
approaching negative three. We could say this is the limit as h of x is approaching negative three, and it is approaching negative three from values greater than negative three. H of x is approaching
negative three from above, or we could say from values
greater than negative three, and then of g of h of x. Another way to think about it, what is the limit as the input to g approaches negative three from the right? As we approach negative
three from the right, g is right here at three, so this is going to be
equal to three again. So notice the right hand
limit and the left hand limit in this case are both equal to three. So when the right hand
and the left hand limit is equal to the same thing,
we know that the limit is equal to that thing. This is a pretty cool example, because the limit of, you
could say the internal function right over here of h of x, did not exist, but the limit of the composite
function still exists.