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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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# Theorem for limits of composite functions: when conditions aren't met

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). Created by Sal Khan.

## Want to join the conversation?

- Can someone please help verify my work? I want to show that f(g(x)) as x->2 does not exist.

*If...*

lim(g(x)) as x->2+ = 0.

from the graph of g, the values of g(x) are approaching 0 from "above" (This will be the input into f)*see video @*1:34

so if we approach 0 from "above" in f(x) we can see that the function f(x) is approaching 3 from the right. This is another way of saying lim f(x) as x->0+ = 3

*And if...*

lim(g(x)) as x->2- is -2.

from the graph of g, the values of g(x) are approaching -2 from "above" (This will be the input into f) So we need to do lim f(x) as x->-2+ = 2

*This means that ...*

because the two limits are not the same (2 and 3) f(g(x))'s limit does not exist.

Please let me know if I'm correct/where I messed up!(40 votes)- I got the same exact thing I think you're absolutely correct =)(13 votes)

- 1:36why does he say "approaching 2 from above" when he is approaching from below on the f(x) graph, and vice versa a few seconds later?(18 votes)
- I suppose g(x) to be another variable, for example g(x)=X. So lim f(g(x)) become lim f(X). As the x in g(x) graph approaches 0 from the left, g(x) approaches 2 from above, which means X in f(X) graph approaches 2 from the bigger side (the right side). When X approaches 2 from the right, its graph f(X) (or f(g(x)) as I've supposed before) approaches 0 from below

Hope this help!(6 votes)

- So basically what I understood from the video is, if the first condition is true (i.e., limit of
`g(x)`

as`x-->a`

exists = L) but the second (`f(x)`

being continuous at L) isn't, then I gotta find the**limit**of f(x) as x-->L. And if the first condition is not true, then the limit**wouldn't exist**at all, because there's no L to begin with.

Have I gotten it right? Any help would be appreciated.(5 votes)- I watched the next two videos on the series and found out that my understanding was wrong. Here are the videos:

https://www.khanacademy.org/math/calculus-1/cs1-limits-and-continuity/cs1-properties-of-limits/v/limits-of-composite-functions-internal-limit-doesn-t-exist

https://www.khanacademy.org/math/calculus-1/cs1-limits-and-continuity/cs1-properties-of-limits/v/limits-of-composite-functions-external-limit-doesn-t-exist

I'm not deleting this question, cuz if anyone had the same doubt (I highly doubt it cuz it was pretty stupid of me lol) they can just watch these videos and hope they clear out things.

If anyone else has a better explanation than the above videos, please do post it here.

Ciao :)(16 votes)

- does anyone know where i can find a proof of this theorem?(8 votes)
- the constantly changing terminology does not help with the explanation. Switching from to the right of, or to the left of; to above or below is very confusing. Anyone have a better explanation?(8 votes)
- lim f(g(x))x->2 does not exist because f(-2)=2, and f(0)=3. If f(-2) = f(0), there would be a limit. Is this right?(4 votes)
- I think you are correct according to the next video 🤔

if lim f(-2) when approaching from positive side = lim f(0) when approaching from positive side then there would be a limit(4 votes)

- What's the point of the "theorem" if you can bypass it? Also, what's the theorem he's talking about called?(4 votes)
- A theorem is just a proven statement that says "if these conditions are met, then this result follows." Sal is demonstrating the converse: if these conditions aren't met, then this result doesn't follow. That doesn't invalidate the theorem.

Also, the theorem he's talking about doesn't have a common name like "Rolle's theorem" or "the Pythagorean theorem" do.(4 votes)

- I did not quite understand from the analysis part.

Please help...(4 votes) - if you can skip conditions, why would you set the conditions in the first place?(2 votes)
- The conditions in the theorem are set to ensure that the theorem can be applied accurately and that the limit of the composite function can be determined correctly. If the conditions of the theorem are met, it provides a straightforward way to evaluate the limit of a composite function by first evaluating the limits of the individual component functions and then applying the function's composition.

However, in some cases, the conditions of the theorem may not be met, and the theorem cannot be directly applied. In such situations, further analysis and reasoning may be required to determine the limit of the composite function. Sal, in the video example, illustrates how he uses his understanding of the behavior of f and g from both the left and right sides of the limit point to deduce the limit of the composite function even though the conditions of the theorem are not met.

So, while the conditions in the theorem provide a convenient and efficient method for evaluating limits of composite functions in certain cases, they may not always be applicable, and additional analysis or reasoning may be needed to determine the limit accurately.(5 votes)

- How can you determine if the function is continuous?(2 votes)
- A function is continuous if every input has a defined output. All polynomials are continuous. But, a lot of rational functions aren't (Ex: x/(x-2) is discontinuous at x = 2).(4 votes)

## Video transcript

- [Tutor] In a previous
video we used this theorem to evaluate certain types
of composite functions. In this video we'll do
a few more examples, that get a little bit more involved. So let's say we wanted
to figure out the limit as x approaches zero of f of g of x, f of g of x. First of all, pause this video and think about whether
this theorem even applies. Well, the first thing to think about is what is the limit as x
approaches zero of g of x to see if we meet this first condition. So if we look at g of x, right over here as x approaches zero from the left, it looks like g is approaching two, as x approaches zero from the right, it looks like g is approaching two and so it looks like this
is going to be equal to two. So that's a check. Now let's see the second condition, is f continuous at that limit at two. So when x is equal to two, it does not look like f is continuous. So we do not meet this second
condition right over here, so we can't just directly
apply this theorem. But just because you
can't apply the theorem does not mean that the limit
doesn't necessarily exist. For example, in this situation the limit actually does exist. One way to think about it, when x approaches zero from the left, it looks like g is
approaching two from above and so that's going to be the input into f and so if we are now
approaching two from above here as the input into f, it looks like our function
is approaching zero and then we can go the other way. If we are approaching zero from
the right, right over here, it looks like the value of our function is approaching two from below. Now if we approach two from below, it looks like the value
of f is approaching zero. So in both of these scenarios, our value of our function
f is approaching zero. So I wasn't able to use this theorem, but I am able to figure out that this is going to be equal to zero. Now let me give you another example. Let's say we wanted to
figure out the limit as x approaches two of f of g of x. Pause this video, we'll first see if this
theorem even applies. Well, we first wanna see what is the limit as x approaches two of g of x. When we look at approaching
two from the left, it looks like g is
approaching negative two. When we approach x equals
two from the right, it looks like g is approaching zero. So our right and left hand
limits are not the same here, so this thing does not
exist, does not exist and so we don't meet this
condition right over here, so we can't apply the theorem. But as we've already seen, just because you can't apply the theorem does not mean that the
limit does not exist. But if you like pondering things, I encourage you to see that
this limit doesn't exist by doing very similar analysis to the one that I did
for our first example.