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L'Hopital's Rule Explained in German


Date: 12/10/1999 at 10:11:07
From: Janine
Subject: Grenzwert einer Folge

Can you tell me in German what the "Grenzwert" of the following 
"Folge" is?

     2^(n+1)-3 / 2^(n-1)

Thanks,
Janine


Date: 12/10/1999 at 10:32:33
From: Doctor Allan
Subject: Re: Grenzwert einer Folge

Hallo Janine,

Kennst du L'Hopitals regel? Er sagt dass wenn

     lim f(x) = lim g(x) = unendlich,

dann ist

     lim (f(x)/g(x)) = lim f'(x)/g'(x)

Deine situation ist die folgende:

     f(n) = 2^(n+1)-3  =>  lim f(x) = unendlich
     g(n) = 2^(n-1)    =>  lim g(x) = unendlich

Benutze L'Hospital:

     f'(n) = 2^(n+1)*ln(2)
     g'(n) = 2^(n-1)*ln(2)

Das heisst

     f'(n)/g'(n) = (2^(n+1)*ln(2))/(2^(n-1)*ln(2))

                 = (2^(n+1))/(2^(n-1))

                 = ((2^2)*(2^(n-1)))/(2^(n-1))

                 = 2^2 * (2^(n-1))/(2^(n-1))

                 = 2^2 = 4

Daher ist lim f'(n)/g'(n) = 4, so dass lim (f(n)/g(n)) = 4. 
Hoffentlich hilft es dir.

Viele grusse,
- Doctor Allan, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Calculus

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