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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
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  1. CliffSedge Group Title
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    A limit of 0 when?

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

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

    • 2 years ago
  4. sevenshaded Group Title
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    3^n

    • 2 years ago
  5. CliffSedge Group Title
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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 ∞/∞.

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

    • 2 years ago
  7. sevenshaded Group Title
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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?

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

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

    • 2 years ago
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