What is the limit of (1 - 1/n)^n as n approaches infinity?

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The limit of (1 - 1/n)^n as n approaches infinity is definitively equal to 1/e. This foundational result allows for further manipulation, such as evaluating the limit of (1 - 1/n)^{n ln(2)}, which simplifies to (1/e)^{ln(2)}. Consequently, this limit equals 1/2, demonstrating the application of exponential limits in calculus. The discussion confirms the legitimacy of these transformations and their outcomes.

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Homework Statement



Screen_shot_2012_04_04_at_9_13_04_AM.png


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]
 
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Looks good to me.
 
TranscendArcu said:

Homework Statement



Screen_shot_2012_04_04_at_9_13_04_AM.png


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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