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'hopitals 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'hopitals 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.