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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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