Can someone explain why

An = (2^n)/[3^(n+1)]

has a limit of 0. I'm missing something algebraically, I guess.

Question

Can someone explain why 

An = (2^n)/[3^(n+1)] 

has a limit of 0.  I'm missing something algebraically, I guess.

anonymous · Answer

A limit of 0 when?

anonymous · Answer

as n goes to infinity?

anonymous · Answer

What gets bigger faster, 2^n or 3^n?

anonymous · Answer

3^n

anonymous · Answer

So you'll get a number very very very large on the bottom sooner than you get one on the top, so it'll approach (something)/∞ before ∞/(something) or ∞/∞.

anonymous · Answer

There are more rigorous mathematical ways to show that, but do you understand the reasoning?

anonymous · Answer

I see.   I mean, I understand it conceptually.  Just not sure how I "show" that when I'm writing things out.    "infinity/infinity" for work shown?

anonymous · Answer

And if it is infinity over infinity, then how would that converge to 0?

anonymous · Answer

Are you familiar with calculus derivatives and L'Hopital's rule?