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Re: Problem with limiits

• To: mathgroup at smc.vnet.net
• Subject: [mg65845] Re: Problem with limiits
• From: Roger Bagula <rlbagulatftn at yahoo.com>
• Date: Tue, 18 Apr 2006 06:56:40 -0400 (EDT)
• References: <200604160544.BAA07913@smc.vnet.net> <e1vdsq$967$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

Andrzej Kozlowski wrote:

>>
> 
> 
> 
> For any sequence of real numbers a1,a2, a3 .... converging to a real  
> number L,  any infinite subsequence of it will converge to the same  
> limit L. This is a very basic fact and is in fact true in any  
> complete metric space (a subsequence of a convergent sequence is a  
> Cauchy sequence, and therefore, in a complete metric space, is itself  
> convergent.)
> 
> Given the above,  you only need the fact that there are infinitely  
> many primes (proved by Euclid and before him by Eudoxus).
> 
> Andrzej Kozlowski
> 
What you are saying is that if I have a composite number c[m]
them the Limit:

Limit[(1+1/c[m])^c[m],m->Infinity]=E

as well.
What About:
Limit[(1-1/m)^m,m->Infinity] =1/E

Your reasoning gives:
Limit[(1-1/c[n])^c[n],n->Infinity]*Limit[(1+1/Prime[n])^Prime[n],N->Infinity]=1
and
Limit[(1-1/m)^m,m->Infinity]*Limit[(1+1/m)^m,m->Infinity]=1
This does give:
Limit[(1 - 1/m^2)^m, m -> Infinity]=1

Thanks for your help.

• Follow-Ups:
  • Re: Re: Problem with limiits
    • From: Andrzej Kozlowski <akoz@mimuw.edu.pl>

• References:
  • Problem with limiits
    • From: Roger Bagula <rlbagulatftn@yahoo.com>