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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 combined functions: piecewise functions

AP.CALC:

LIM‑1 (EU)

, LIM‑1.D (LO)

, LIM‑1.D.1 (EK)

, LIM‑1.D.2 (EK)

Even when the limits of two functions at some point do not exist, the limit of their sum or product might still exist.

## Want to join the conversation?

- In this video, it seems that main idea interferes with the limit properties video previously, and the traditional view of limits learned through most calculus texts. The general idea is the limit of sum is the same as the sum of the limits... If we then take the sum of each limit, each limit is DNE so how can we combine them in the traditional sense?(36 votes)
- The idea about the existence of the limit of a function at any value "p" is that the one sided limits as x -> p are equal. If we make the graph of the combined functions showed in the video we will see that the one sided limits are equal in the first and third case but not in the second. There will be a discontinuity when the limit doesn't exist.(16 votes)

- i can pass the test about this video but i will never understand the proof of this.can s1 explain?(10 votes)
- By combining f(x) and g(x) you are basically creating a new function that does not necessarily follow the same principles as the original functions.

Take for example:

f(x) = 1/(x-1)

g(x) = (x-1)^2

h(x) = f(x)*g(x) = ((x-1)(x-1))/(x-1) = x-1

If we wanted to calculate the limit of x->1 of f(x) we would get:

1/1-1 = 1/0, which is undefined

However, if we take this same limit from h(x) we would simply get 0, as the new function is x-1(1 vote)

- i cant understand if the limits dont exist how can we sum them ?(8 votes)
- We are finding the limit of a complex function, not the sum of two individual limits. If you add the two functions together you'll find out that the function is actually continuous.(2 votes)

- So, in short, it doesn't matter if the individual limits exist, if the left and right of lim(f(x) + g (x) ) equal, there is a limit?(7 votes)
- The individual limit does exist. Take x -> -2 (f(x) + g(x)) for example. Think of (f(x) + g(x)) as a single function that can be represented by f(x) and g(x). If you combine them, you will realize both the limits approaching from the right and left are 4. So in general, view whatever inside the parenthesis as a single function THEN take the limit.(3 votes)

- When dealing with combined functions, why do we need to check the limit as it approaches from the left side and right side?

Whereas in the later on composite function video, we do not have to check the limit as it approaches from the left side and right side? It makes sense, however I am having difficulty putting it into words.(5 votes)- For a limit to exist, it has to a approach a certain value from both the right and left side, and when you have the limit of the sum or product of two functions, even if the limit of each function doesn't exist, the limit of the sum/product might still exist if it meets this condition.

Also, in the later video, Sal**is**taking the limit from the right and left side, if from both sides it approached a different value the limit wouldn't exist.(6 votes)

- At minute0:30, as x approaches -2 from the positive side, doesn't the y value approach 3? The video says that the y value approaches 2 when x approaches -2 from the positive side. Is this a mistake, or am I missing something? Many thanks,(1 vote)
- i think he corrected the video because it is correct as of now (which is three months aafter this comment)(4 votes)

- This video was so confusing just because Sal mixed up left and right so many times. If you're approaching FROM the left-hand side, you're going TOWARDS the right-hand side. From the left to the right. From the right to the left.

Sal went from the right to the right and from the left to the left, which didn't make any sense. This video would be helpful with some edits, or even just doing the video over again.(1 vote)- Although it is confusing, just think of it as literally looking at the left hand side of the x value you're taking the limit of or taking the right. While this isn't the perfect terminology (it should be "approaching," of course), this might be easier to remember! :)(3 votes)

- Why do you need to add up f(x) and g(x) right and left approach individually? It looks like Sal is just looking for the same approach of the composite's left function's left approach, with the composite's right function's right approach. Will this always be true?(2 votes)
- if you graph the final function (e.g. f(x)+g(x)), you would see the 4 different dots of 2 functions (including empty ones) converging to 1 value (empty or not)

thus it doesn't matter if you approach this from left or right at the final stage. but we did tackle them individually, when it's hard to graph ourselves, and it is easier to simply calculate them manually. then it does matter don't mix up the direction of approach (just as the graphing machine doesn't)(1 vote)

- so what if finding the limit from the right is different from finding the left, does it also stand that the limit does not exist(1 vote)
- Yep. If the two sided limits do not approach the same value, the limit doesn't exist.(2 votes)

- Can the limit ever exist if one function is continuous while the other isn't? I'm thinking if you had the same number for both the left approach and the right approach (from the continuous function) while having different numbers added (from the piece wise) you'd have different numbers for the left and right approach since you'd be adding a different number to the same number so you'd always come out with DNE for a couple of functions where one is continuous and one isn't?(1 vote)
- That's a really good question! Had me thinking for a solid 1 hour (mostly because I couldn't find more than one example to disprove it lol!)

There is a special case where if one function is continuous and the other isn't, the sum of limits will still exist. This only happens (as far as I know) with piecewise functions with point discontinuities. For example, if I have a continuous function f(x) and a discontinuous function g(x) which has a point discontinuity at x = 3, the limit as (f(x) + g(x)) tends to 3 will still exist, as the only difference between f(x) and g(x) is that in g(x), x=3 isn't in its domain.

In general though, the limit will indeed not exist, as, just like you said, you're adding/subtracting different numbers from the discontinuous function to the same number in the continuous one, hence yielding different numbers and thus, a non-existent limit.(1 vote)

## Video transcript

- [Instructor] We are asked to find these three different limits. I encourage you like always, pause this video and try to do it yourself before we do it together. So when you do this first one, you might just try to find the limit as x approaches negative two of f of x and then the limit as x
approaches negative two of g of x and then add those two limits together. But you will quickly find a problem, 'cause when you find the limit as x approaches negative two of f of x, it looks as we are
approaching negative two from the left, it looks
like we're approaching one. As we approach x equals
negative two from the right, it looks like we're approaching three. So it looks like the limit as x approaches negative
two of f of x doesn't exist, and the same thing's true of g of x. If we approach from the left, it looks like we're approaching three. If we approach from the right, it looks like we're approaching one. But it turns out that
this limit can still exist as long as the limit as
x approaches negative two from the left of the sum, f of x plus g of x, exists and is equal to the limit as x approaches negative two
from the right of the sum, f of x plus g of x. So what are these things? Well, as we approach
negative two from the left, f of x is approaching, looks like one, and g of x is approaching three. So it looks like we're
approaching one and three. So it looks like this is approaching. The sum is going to approach four. And if we're coming from the right, f of x looks like it's approaching three and g of x looks like
it is approaching one. Once again, this is equal to four. And since the left and right handed limits are approaching the same thing, we would say that this limit
exists and it is equal to four. Now let's do this next
example as x approaches one. Well, we'll do the exact same exercise. And once again, if you look
at the individual limits for f of x from the left and
the right as we approach one, this limit doesn't exist. But the limit as x approaches
one of the sum might exist, so let's try that out. So the limit as x approaches one from the left hand side
of f of x plus g of x, what is that going to be equal to? So f of x, as we approach
one from the left, looks like this is approaching two. I'm just doing this for shorthand. And g of x, as we approach
one from the left, it looks like it is approaching zero. So this will be approaching
two plus zero, which is two. And then the limit, as x approaches one
from the right hand side of f of x plus g of x
is going to be equal to. Well, for f of x as we're approaching one from the right hand side, looks like it's approaching negative one. And for g of x as we're approaching one from the the right hand side, looks like we're approaching zero again. Here it looks like we're
approaching negative one. So the left and right hand limits aren't approaching the same value, so this one does not exist. And then last but not least, x approaches one of f of x times g of x. So we'll do the same drill. Limit as x approaches one
from the left hand side of f of x times g of x. Well, here, and we can
even use the values here. We see it was approaching
one from the left. We are approaching two, so this is two. And when we're approaching
one from the left here, we're approaching zero. We're gonna be approaching
two times zero, which is zero. And then we approach from the right. X approaches one from the right of f of x times g of x. Well, we already saw when
we're approaching one from the right of f of x, we're approaching negative one. But g of x, approaching
one from the right, is still approaching zero, so this is going to be zero
again, so this limit exists. We get the same limit when we approach from
the left and the right. It is equal to zero. So these are pretty interesting examples, because sometimes when you think that the component limits don't exist that that means that the sum or the product might not exist, but this shows at least two examples where that is not the case.