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sevenshaded

  • 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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  1. CliffSedge
    • 3 years ago
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    A limit of 0 when?

  2. sevenshaded
    • 3 years ago
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    as n goes to infinity?

  3. CliffSedge
    • 3 years ago
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    What gets bigger faster, 2^n or 3^n?

  4. sevenshaded
    • 3 years ago
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    3^n

  5. CliffSedge
    • 3 years ago
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    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 ∞/∞.

  6. CliffSedge
    • 3 years ago
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    There are more rigorous mathematical ways to show that, but do you understand the reasoning?

  7. sevenshaded
    • 3 years ago
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    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?

  8. sevenshaded
    • 3 years ago
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    And if it is infinity over infinity, then how would that converge to 0?

  9. CliffSedge
    • 3 years ago
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    Are you familiar with calculus derivatives and L'Hopital's rule?

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