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# Limits of combined functions: piecewise functions

This video demonstrates that even when individual limits of functions f(x) and g(x) don't exist, the limit of their sum or product might still exist. By analyzing left and right-hand limits, we can determine if the limit of the combined functions exists and find its value.

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• 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?
• 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.
• i can pass the test about this video but i will never understand the proof of this.can s1 explain?
• 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
• i cant understand if the limits dont exist how can we sum them ?
• 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.
• 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?
• 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.
• 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.
• 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.
• 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?
• 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.
• At minute , 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)
• With respect to the first example, the question was straightforward, and the limit was undefined. Now, must we always prove the limit exists even if the question does not ask that?
(1 vote)
• well if ur taking the ap exam and ur doing one of the free response questions, then yeah. u always gotta analyze the limit from the left and right and then, seeing their the same,declare the limit exists.
• why does dividing with 0 result in DNE but multiplying by 0 does not?
(1 vote)
• Dividing any number by 0 is always undefined, but multiplying by 0 is always 0. So the limit only doesn't exist when Dividing by 0 because multiplying by 0 still results in a defined answer.