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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0as n goes to infinity?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So 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
 3 years ago
Best ResponseYou've already chosen the best response.0There are more rigorous mathematical ways to show that, but do you understand the reasoning?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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
 3 years ago
Best ResponseYou've already chosen the best response.0And if it is infinity over infinity, then how would that converge to 0?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Are you familiar with calculus derivatives and L'Hopital's rule?
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