
#1
Feb1708, 09:35 PM

P: 2,268

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




#2
Feb1708, 09:42 PM

P: 813





#3
Feb1708, 09:54 PM

P: 2,268

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



#4
Feb1708, 10:09 PM

P: 813

Proving the existence of a limit?http://en.wikipedia.org/wiki/Absolute_convergence 



#5
Feb1708, 10:51 PM

P: 2,268

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?




#6
Feb1708, 10:54 PM

P: 813





#7
Feb1708, 11:00 PM

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#8
Feb1708, 11:03 PM

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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. 



#9
Feb1708, 11:18 PM

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#10
Feb1708, 11:34 PM

P: 2,268





#11
Feb1708, 11:41 PM

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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, [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] 



#12
Feb1708, 11:50 PM

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#13
Feb1808, 12:01 AM

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#14
Feb1808, 12:26 AM

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#15
Feb1808, 12:41 AM

P: 2,268

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? 



#16
Feb1808, 01:07 AM

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#17
Feb1808, 06:33 AM

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Why didn't you say that when you were first asked to be specific?
If all you know is that [itex]a_n[/itex] and [itex]b_n[/itex] go to 0, you can't say anything about whether [itex]a_n/b_n[/itex] converges or diverges. For example, if [itex]a_n= 1/n^2[/itex] and [itex]b_n= 1/n[/itex], it is obvious that both [itex]a_n[/itex] and [itex]b_n[/itex] converge to 0. And [itex]a_n/b_n= (1/n^2)(n/1)= 1/n[/itex] converges to 0. But if [itex]b_n= 1/n^2[/itex] and [itex]a_n= 1/n[/itex], it is still obvious that both [itex]a_n[/itex] and [itex]b_n[/itex] converge to 0 but now [itex]a_n/b_n= (1/n)(n^2/1)= n[/itex] does not converge. 



#18
Feb2008, 12:12 PM

P: 239

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



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