Lim 1/2^x-1 as x goes to infinity.

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SUMMARY

The limit of the expression 1/2^x - 1 as x approaches infinity is -1. The discussion clarifies that applying L'Hôpital's rule is unnecessary because the numerator remains constant at 1 while the denominator approaches infinity. Therefore, the limit simplifies directly to -1 without further complex calculations. The correct interpretation of the expression is crucial for accurate evaluation.

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I can not understand what to do with this

Lim 1/2^x-1 as x goes to infinity.

i tried using l'hospital's rule but it does't work or I am not applying it right.
 
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You should include () to clarify.
(1/2^x)-1 or 1/2^(x-1)
 
2slowtogofast said:
I can not understand what to do with this

Lim 1/2^x-1 as x goes to infinity.

i tried using l'hospital's rule but it does't work or I am not applying it right.
[itex](1/2)^x- 1[/itex] goes to -1, of course,
[itex]1/(2^{x-1})[/itex] goes to 0.

[tex]\frac{1}{2^x- 1}[/tex] also goes to 0. You don't need L'Hopital because the numerator does not go to infinity. The numerator is always 1 while the denominator goes to infinity.
 

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