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

This Question is Closed

CliffSedge Group TitleBest ResponseYou've already chosen the best response.0
A limit of 0 when?
 2 years ago

sevenshaded Group TitleBest ResponseYou've already chosen the best response.0
as n goes to infinity?
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.0
What gets bigger faster, 2^n or 3^n?
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.0
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 ∞/∞.
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.0
There are more rigorous mathematical ways to show that, but do you understand the reasoning?
 2 years ago

sevenshaded Group TitleBest ResponseYou've already chosen the best response.0
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?
 2 years ago

sevenshaded Group TitleBest ResponseYou've already chosen the best response.0
And if it is infinity over infinity, then how would that converge to 0?
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.0
Are you familiar with calculus derivatives and L'Hopital's rule?
 2 years ago
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