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

[2slowtogofast]

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.

 

[mathman]

You should include () to clarify.
(1/2^x)-1 or 1/2^(x-1)

 

[HallsofIvy]

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.

$(1/2)^x- 1$ goes to -1, of course,
$1/(2^{x-1})$ goes to 0.

$$\frac{1}{2^x- 1}$$ 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.