# Math Help - How would you about calculating this simple limit?

1. ## How would you go about calculating this simple limit?

a(n) = (2^(n-1)) / ((2^n)-1)
with n --> infinite.

thank you

2. ## Re: How would you about calculating this simple limit?

Originally Posted by nappysnake
a(n) = (2^(n-1)) / ((2^n)-1)
with n --> infinite.
$\lim_{n \to \infty} \frac{2^{n-1}}{2^n - 1} =$

divide every term by $2^{n-1}$ ...

$\lim_{n \to \infty} \frac{1}{2 - \frac{1}{2^{n-1}}} = \frac{1}{2}$

3. ## Re: How would you about calculating this simple limit?

thank you skeeter for letting me come to a crucial realization, it was very clear