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-   -   Limits as n goes to infinity (http://www.physicsforums.com/showthread.php?t=593379)

TranscendArcu Apr4-12 08:22 AM

Limits as n goes to infinity
 
1. The problem statement, all variables and given/known data

http://s14.postimage.org/v6tcx598h/S...9_13_04_AM.png

3. The attempt at a solution
So I know that the limit as [itex]n → ∞[/itex] of [itex](1 - \frac{1}{n})^n = \frac{1}{e}[/itex]. Using this information, is it legitimate to observe:

The limit as [itex]n → ∞[/itex] of [itex](1 - \frac{1}{n})^{n ln(2)} =[/itex] the limit as [itex]n → ∞[/itex] of [itex]((1 - \frac{1}{n})^n)^{ln(2)} = e^{-1 ln(2)} = e^{ln(\frac{1}{2})} = \frac{1}{2}[/itex]

Robert1986 Apr4-12 08:48 AM

Re: Limits as n goes to infinity
 
Looks good to me.

Dick Apr4-12 08:49 AM

Re: Limits as n goes to infinity
 
Quote:

Quote by TranscendArcu (Post 3849552)
1. The problem statement, all variables and given/known data

http://s14.postimage.org/v6tcx598h/S...9_13_04_AM.png

3. The attempt at a solution
So I know that the limit as [itex]n → ∞[/itex] of [itex](1 - \frac{1}{n})^n = \frac{1}{e}[/itex]. Using this information, is it legitimate to observe:

The limit as [itex]n → ∞[/itex] of [itex](1 - \frac{1}{n})^{n ln(2)} =[/itex] the limit as [itex]n → ∞[/itex] of [itex]((1 - \frac{1}{n})^n)^{ln(2)} = e^{-1 ln(2)} = e^{ln(\frac{1}{2})} = \frac{1}{2}[/itex]

Looks ok to me.


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