- #1

- 4

- 0

I was just wondering how to solve this limit: Limit of (3/4)^(n+1) as n approaches infinity

My attempt:

(3/4)^(n+1) = (3^ (n+1) ) / (4 ^ (n+1))

Top goes to infinity and bottom goes to infinity to use l'hopital rule.

lim = ( ln(3) * 1 * 3^(n+1) ) / ( ln(4) * 1 * 4^(n+1) )

But the correct answer is apparently 0.

Not sure if i'm totally wrong or if I'm just halfway there.

Thanks