# Proving the existence of a limit?

## Main Question or Discussion Point

How do you prove the existence but not necessarily the value of a limit?

How do you prove the existence but not necessarily the value of a limit?
Why not give an example of such a problem because I think your question is too general.

Suppose you are given a limit. Show that a limit exists without needing to say what it is.

It is a general question.

Okay, good point. I'll be more specific. I was thinking of limits of sequences only. Or are all limits sequences in which case I haven't narrowed down anything?

Okay, good point. I'll be more specific. I was thinking of limits of sequences only. Or are all limits sequences in which case I haven't narrowed down anything?
Well how about showing the limit of $$a_n-a_{n+1}=0$$ as n->oo

morphism
Homework Helper
Well how about showing the limit of $$a_n-a_{n+1}=0$$ as n->oo
This doesn't imply that (a_n) will converge. [Counterexample: a_n = log(n).]

morphism
Homework Helper
Okay, good point. I'll be more specific. I was thinking of limits of sequences only. Or are all limits sequences in which case I haven't narrowed down anything?
If you're going to be taking a limit of something, then it had better be a sequence (or something sequence-like). So, no, you haven't narrowed down anything. Do you mean sequences of numbers?

In any case, there are a few existence results that can come in handy. One example is the monotone convergence theorem, which states that if a sequences of real numbers is monotone and bounded, then it must converge.

Well how about showing the limit of $$a_n-a_{n+1}=0$$ as n->oo
This doesn't imply that (a_n) will converge. [Counterexample: a_n = log(n).]
What if a_n is bound both above and bellow?

What if a_n is bound both above and bellow?
In which case it would be in a compact space. All cauchy sequences in a compact space converge.

morphism
Homework Helper
In which case it would be in a compact space. All cauchy sequences in a compact space converge.
Just because (a_n - a_{n+1}) -> 0 doesn't mean a_n is Cauchy. Again, a counterexample to this claim is a_n = log(n). We have log(n) - log(n+1) = log(n/(n+1)) -> log(1) = 0, but if this were Cauchy, then it would be convergent (since the reals are complete), and it clearly isn't.

It turns out that being bounded isn't good enough either, although finding a counterexample was trickier. At any rate: try a_n = exp(i(1 + 1/2 + ... + 1/n)). This doesn't converge - it goes around the unit circle in the complex plane. On the other hand,
$$a_n - a_{n+1} = \exp\left(i \left(1 + \frac{1}{2} + ... + \frac{1}{n}\right)\right)\left(1 - \exp\left(\frac{i}{n+1}\right)\right) \to 0.[/itex] Last edited: Just because (a_n - a_{n+1}) -> 0 doesn't mean a_n is Cauchy. Again, a counterexample to this claim is a_n = log(n). We have log(n) - log(n+1) = log(n/(n+1)) -> log(1) = 0, but if this were Cauchy, then it would be convergent (since the reals are complete), and it clearly isn't. It turns out that being bounded isn't good enough either, although finding a counterexample was trickier. At any rate: try a_n = exp(i(1 + 1/2 + ... + 1/n)). This doesn't converge - it goes around the unit circle in the complex plane. On the other hand, [tex]a_n - a_{n+1} = \exp\left(i \left(1 + \frac{1}{2} + ... + \frac{1}{n}\right)\right)\left(1 - \exp\left(\frac{i}{n+1}\right)\right) \to 0.[/itex] Okay, how about [tex]a_n-a_{n-1} \le k \left( \frac{1}{n}-\frac{1}{n-1} \right)$$ for all n greater then M

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It turns out that being bounded isn't good enough either, although finding a counterexample was trickier. At any rate: try a_n = exp(i(1 + 1/2 + ... + 1/n)). This doesn't converge - it goes around the unit circle in the complex plane. On the other hand,
$$a_n - a_{n+1} = \exp\left(i \left(1 + \frac{1}{2} + ... + \frac{1}{n}\right)\right)\left(1 - \exp\left(\frac{i}{n+1}\right)\right) \to 0.[/itex] That's a pretty interesting example. Can you find a similar example on the reals? morphism Science Advisor Homework Helper Okay, how about [tex]a_n-a_{n-1} \le k \left( \frac{1}{n}-\frac{1}{n-1} \right)$$ for all n greater then M
What are k and M?

That's a pretty interesting example. Can you find a similar example on the reals?
Take the real part of it and see what happens.

To be more specific, I had a sequence a_m/b_m where a_m and b_m in the limit is 0.

However they are not functions so can't use that rule which I can't spell, l'hospil's rule?

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What are k and M?

Take the real part of it and see what happens.
k an M are arbitrary chosen to make the inequality work for a given sequence.

HallsofIvy
Homework Helper
Why didn't you say that when you were first asked to be specific?

If all you know is that $a_n$ and $b_n$ go to 0, you can't say anything about whether $a_n/b_n$ converges or diverges. For example, if $a_n= 1/n^2$ and $b_n= 1/n$, it is obvious that both $a_n$ and $b_n$ converge to 0. And $a_n/b_n= (1/n^2)(n/1)= 1/n$ converges to 0.

But if $b_n= 1/n^2$ and $a_n= 1/n$, it is still obvious that both $a_n$ and $b_n$ converge to 0 but now $a_n/b_n= (1/n)(n^2/1)= n$ does not converge.

ssd
Looking upon the 1st post: if left hand and right hand limits are equal, then the limit exists.