# Algebra of Limits

The theorem that allows one to distribute the limit over addition is the following: Let ##(a_n), (b_n)## be sequences that converge to ##L## and ##M## respectively. Then ##\lim (a_n+b_n) = L + M##.

So evidently, a hypothesis of distributing the limit is that we know ##a_n## and ##b_n## converge.

So, here is my question. Say that I don't know whether ##1/n## and ##1/n^2## converges or not. Normally, to evaluate ##\lim (1/n + 1/n^2)## we distribute the limit and then determine whether each sequence converges or not. Shouldn't this be the other way around? Shouldn't we determine whether each sequence converges first, and then distribute the limit, which is what models the logical progression of the theorem above?

## Answers and Replies

FactChecker
Gold Member
As long as you verify that the two individual limits exist, that shows that the equality holds. You are not using the equality before it is proven except to guide you in identifying the individual limits that you need to prove. You are not using it to tell you that the individual limits do exist.

Mr Davis 97
fresh_42
Mentor
So evidently, a hypothesis of distributing the limit is that we know ##a_n## and ##b_n## converge.
This doesn't work in this direction. First we have to make sure that the components converge, then the formula holds.
An easy counterexample: ##a_n=(-1)^n\; , \;b_n=(-1)^{n+1}##.

WWGD
WWGD
Gold Member
This doesn't work in this direction. First we have to make sure that the components converge, then the formula holds.
An easy counterexample: ##a_n=(-1)^n\; , \;b_n=(-1)^{n+1}##.
Or, in the opposite direction ##a_n =2^n ## and ## b_n =-2^n ##

Last edited:
Stephen Tashi
Normally, to evaluate ##\lim (1/n + 1/n^2)## we distribute the limit and then determine whether each sequence converges or not. Shouldn't this be the other way around? Shouldn't we determine whether each sequence converges first, and then distribute the limit, which is what models the logical progression of the theorem above?
Yes, to be perfectly rigorous we should do that. However, as you said, it's common to see a sequence of equations like
##lim_{n \rightarrow \infty } ((f(n) + g(n)) = \lim_{n \rightarrow \infty} f(n) + \lim_{n \rightarrow \infty} g(n) = L_1 + L_2 ##
This turns out to be true "in the end" when ## \lim_{n \rightarrow \infty} f(n) = L_1## and ##\lim_{n \rightarrow \infty} g(n) = L_2## since both limits exist. However, the rigorous way to write things would be to write that both those limits exist before stating the value of ##\lim_{n \rightarrow \infty} (f(n) + g(n))##.

As pointed out by others, if you are trying to prove ##lim_{n \rightarrow \infty} ( f(n) + g(n)) ## does not exist, then it isn't sufficient to show that one or both of ##lim_{n\rightarrow \infty} f(n)## or ##\lim_{n \rightarrow \infty} g(n)## don't exist.

WWGD
Gold Member
Yes, to be perfectly rigorous we should do that. However, as you said, it's common to see a sequence of equations like
##lim_{n \rightarrow \infty } ((f(n) + g(n)) = \lim_{n \rightarrow \infty} f(n) + \lim_{n \rightarrow \infty} g(n) = L_1 + L_2 ##
This turns out to be true "in the end" when ## \lim_{n \rightarrow \infty} f(n) = L_1## and ##\lim_{n \rightarrow \infty} g(n) = L_2## since both limits exist. However, the rigorous way to write things would be to write that both those limits exist before stating the value of ##\lim_{n \rightarrow \infty} (f(n) + g(n))##.

As pointed out by others, if you are trying to prove ##lim_{n \rightarrow \infty} ( f(n) + g(n)) ## does not exist, then it isn't sufficient to show that one or both of ##lim_{n\rightarrow \infty} f(n)## or ##\lim_{n \rightarrow \infty} g(n)## don't exist.
But what do you say about the counterexamples like ## a_n =2^n , b_n =-2^n ##? The limit exists but it is not equal to the sum of the limits?

Stephen Tashi
But what do you say about the counterexamples like ## a_n =2^n , b_n =-2^n ##? The limit exists but it is not equal to the sum of the limits?
If we're talking about ##lim_{n \rightarrow \infty} ## what limits would you sum?

The fact that lim (f + g) can exist when the individual limits do not is not a counterexample to anything I said.

It is not a counter example to the theorem that if the individual limits exist then the limit of their sum exists.

It is not a counter example to the statement that we cannot prove the the limit of the sum does not exist by showing the limits of the summands do not exist.

WWGD
Gold Member
If we're talking about ##lim_{n \rightarrow \infty} ## what limits would you sum?

The fact that lim (f + g) can exist when the individual limits do not is not a counterexample to anything I said.

It is not a counter example to the theorem that if the individual limits exist then the limit of their sum exists.

It is not a counter example to the statement that we cannot prove the the limit of the sum does not exist by showing the limits of the summands do not exist.