# How would you about calculating this simple limit?

• Nov 27th 2011, 12:51 PM
nappysnake
How would you go about calculating this simple limit?
a(n) = (2^(n-1)) / ((2^n)-1)
with n --> infinite.

thank you :)
• Nov 27th 2011, 01:04 PM
skeeter
Re: How would you about calculating this simple limit?
Quote:

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 $2^{n-1}$ ...

$\lim_{n \to \infty} \frac{1}{2 - \frac{1}{2^{n-1}}} = \frac{1}{2}$
• Nov 27th 2011, 01:11 PM
nappysnake
Re: How would you about calculating this simple limit?
thank you skeeter for letting me come to a crucial realization, it was very clear :)