# Limit ?

1. Apr 25, 2010

### 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 im not applying it right.

2. Apr 25, 2010

### mathman

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

3. Apr 26, 2010

### HallsofIvy

$(1/2)^x- 1$ 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.