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MarcLeclair

  • one year ago

Can I figure out if this sequence is converging with the squeeze theorem? Lim n-->inf. (ln n)^2 /n ?

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  1. MarcLeclair
    • one year ago
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    I did this \[0 \le \ln(n)^{2} \le 1/n \] solving for the limit and both side gives 0 therefore it must give 0? :/ or am I doing something wrong

  2. MarcLeclair
    • one year ago
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    ln(n)^2/n * sorry

  3. genius12
    • one year ago
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    I believe he means (ln(n))^2

  4. genius12
    • one year ago
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    Not ln(n)^2

  5. Kainui
    • one year ago
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    Yeah, ln^2(n) is >1 so you can't use that.

  6. genius12
    • one year ago
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    oh ok

  7. MarcLeclair
    • one year ago
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    Wait I thought (ln(n))^2 meant (ln(n)) * (ln(n))

  8. MarcLeclair
    • one year ago
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    Oh yeah I see it it could be greater than one, stupid question :/

  9. Kainui
    • one year ago
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    It does. ln^2(n) is the same as (ln(n))^2. Putting the squared on the function means that. It's so you don't confuse things like sin^2(x) and sin(x^2). But anyways, what I'm saying Marc is that you can't say: ln^2(n)/n≤1/n because it isn't true.

  10. MarcLeclair
    • one year ago
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    Yeah, just noticed it, anyway thanks, basic math issues I guess ^^

  11. Kainui
    • one year ago
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    Yeah lol ok so your left hand side should also be something better than 0 I think, that way you're actually squeezing. ln(n)/n<ln^2(n)/n for the right side I'm still thinking haha.

  12. MarcLeclair
    • one year ago
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    Well I could use l'hospital rule, I just use it so often and my answer key doesn't which gives me wrong answers so I'm afraid to use it now x.x

  13. Kainui
    • one year ago
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    Haha there's no "s" in L'Hopital.

  14. MarcLeclair
    • one year ago
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    Yikes. Anyway thanks hehe!!

  15. genius12
    • one year ago
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    L'hopital's rule would be much easier to apply here than the squeeze theorem:\[\lim_{n \rightarrow \infty}\frac{ \ln(n)^2 }{ n }=2\lim_{n \rightarrow \infty}\frac{ \ln(n) }{ n }=2\lim_{n \rightarrow \infty}\frac{ 1 }{ n }=2*0=0\] @MarcLeclair @Kainui

  16. Kainui
    • one year ago
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    Hmm I agree with you here, but I thought the question was specifically asking to solve this with the squeeze theorem.

  17. genius12
    • one year ago
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    Well you are supposed to use squeeze theorem which he figured already, but l'hopitals would be easier if we were allowed to use it.

  18. Kainui
    • one year ago
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    Haha yeah for sure. Actually we can just use that simple idea to squeeze it between these two which are easy to also find the limit of with l'hopital's! haha ln(n)/n≤ln^2(n)/n≤ln^3(n)/n ?

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