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anonymous

  • 5 years ago

lim n -> infinity (2n+ 2^n) / (3n+3^n)

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  1. anonymous
    • 5 years ago
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    l'hopital's rule.

  2. anonymous
    • 5 years ago
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    no cant use that.....

  3. anonymous
    • 5 years ago
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    is the answer 2/3?

  4. anonymous
    • 5 years ago
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    not sure if its 2/3 or zero?

  5. anonymous
    • 5 years ago
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    i doubt that. i think it should be zero. the first terms are irrelevant since they are linear and the rest is exponential. if we simply erase the first terms you get \[(\frac{2}{3})^n\]which certainly goes to zero. of course this is not a proof.

  6. anonymous
    • 5 years ago
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    The answer is 0I used Mathematica to get the answer. I dont know how to get there though

  7. anonymous
    • 5 years ago
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    yes the wolfram is also showing zero but I dont know how to get there

  8. anonymous
    • 5 years ago
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    hey should I use??? |a(n) - 0 | < epsilon ???

  9. anonymous
    • 5 years ago
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    i still don't know why l'hopital does not apply. take the derivative once, get \[\frac{2+2^nln(2)}{3+3^nln(3)}\]

  10. anonymous
    • 5 years ago
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    I know we can use it but its analysis course, so I have to find the limit without l'hopital rule

  11. anonymous
    • 5 years ago
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    do it again, get \[\frac{2^n (ln(2))^2}{3^n (ln(3))^2}\]

  12. anonymous
    • 5 years ago
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    ooh sorry.

  13. anonymous
    • 5 years ago
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    any ideas you have....??? but now I m sure its 0. so I m half way I think

  14. anonymous
    • 5 years ago
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    i give up :(

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spraguer (Moderator)
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