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chrisplusian Group Title

find the limit of (n^p)/(e^n) as n approaches infinity

  • one year ago
  • one year ago

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  1. ParthKohli Group Title
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    What is \(p\)?

    • one year ago
  2. chrisplusian Group Title
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    P>0 I really dont even understand why they give that information accept that it shows n remains in the numerator

    • one year ago
  3. chrisplusian Group Title
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    I tried to use logarithmic properties to find an answer but it just kept running me in circles

    • one year ago
  4. ParthKohli Group Title
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    \[n^p \times \frac{1}{e^{n}}\]\(e\) is a constant. Oh, and \(n^p \to \infty \) and \(\dfrac{1}{e^n} \to 0\). That can only mean that their product is approaching 0.

    • one year ago
  5. ParthKohli Group Title
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    Am I right? O_O

    • one year ago
  6. terenzreignz Group Title
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    Definitely not the reasoning I'd pick...

    • one year ago
  7. ParthKohli Group Title
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    Knew it.

    • one year ago
  8. experimentX Group Title
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    correct answer but wrong logic ... think about the opposite e^n/n^p and n->infinity

    • one year ago
  9. terenzreignz Group Title
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    I prefer simple cases... \[\Large 2e^n \rightarrow \infty\]\[\Large \frac1{e^n}\rightarrow 0\] \[\Large 2e^n \times \frac1{e^n }\rightarrow 0 \quad \color{red}?\]

    • one year ago
  10. chrisplusian Group Title
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    See i don't understand becasue if you look at it as (N^p)*(1/e^n) then the n^p is infinity and the !/e^n is zero. That gives you infinity times zero which is an indeterminate form.

    • one year ago
  11. DLS Group Title
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    0

    • one year ago
  12. chrisplusian Group Title
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    Sorry (1/e^n)

    • one year ago
  13. terenzreignz Group Title
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    Relax... you can either use L'Hôpital or one of my favourite theorems~ pick one

    • one year ago
  14. ParthKohli Group Title
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    Approaching infinity, not really infinity. I tried to use l'hopital: didn't work. :-|

    • one year ago
  15. terenzreignz Group Title
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    The almighty Squeeze~

    • one year ago
  16. DLS Group Title
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    Isnt the answer 0?

    • one year ago
  17. ParthKohli Group Title
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    ARMAGERD. How could I forget the squeeze

    • one year ago
  18. chrisplusian Group Title
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    Ok I tried lopitals and it did not work nor did logarithmic properties

    • one year ago
  19. ParthKohli Group Title
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    What's the inequality BTW?

    • one year ago
  20. chrisplusian Group Title
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    @DLS the answer is zero

    • one year ago
  21. DLS Group Title
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    Yes,its easy.

    • one year ago
  22. DLS Group Title
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    Use LH

    • one year ago
  23. ParthKohli Group Title
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    @DLS how?

    • one year ago
  24. experimentX Group Title
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    My favouraite technique is \[ a = e^{\log a} \\ n^{p} = e^{p \log n } \]

    • one year ago
  25. terenzreignz Group Title
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    I realise that L'Hôpital may not work after all... at least, not yet... Use this... Let \(\large q = \lceil p\rceil\) such that... \[\Large 0 \le\frac{n^p}{e^n}\le \frac{n^q}{e^n}\]

    • one year ago
  26. terenzreignz Group Title
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    q is guaranteed to be a positive integer, means it [the numerator] will (eventually) differentiate to zero...

    • one year ago
  27. chrisplusian Group Title
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    @experimentX why would that help?

    • one year ago
  28. experimentX Group Title
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    L'hopital works use it least floor(p + 1)

    • one year ago
  29. terenzreignz Group Title
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    Why deviate from the norm, @experimentX ? :D Just use ceiling :3

    • one year ago
  30. DLS Group Title
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    Well observe the denominator. Its e^n.No matter how many times you differentiate it,i.e use L-Hopsital,the denominator WON'T CHANGE at all.That is the first thing.Now I say,use L hospital infinity times(not really) so if you keep on differentiating..a time will come when numerator turns out to be 0. so 0/1=1. Example: \[y=5x^4\] \[\frac{dy}{dx}=20x^3=>60x^2=>120x=>120\]

    • one year ago
  31. chrisplusian Group Title
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    Don't understand what you mean @experimentX

    • one year ago
  32. experimentX Group Title
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    @chrisplusian |dw:1369325101292:dw|

    • one year ago
  33. terenzreignz Group Title
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    To simplify things, we best make sure the exponent p in the numerator is an (positive, but that's granted) integer. Taking \(\Large q = \lceil p\rceil\) helps with that.

    • one year ago
  34. terenzreignz Group Title
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    And then squeezing :)

    • one year ago
  35. experimentX Group Title
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    @terenzreignz yes ceil

    • one year ago
  36. DLS Group Title
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    @chrisplusian my reply should probably clear it off. @experimentX does it seem right?

    • one year ago
  37. chrisplusian Group Title
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    the solutions guide says =0 when p>0, and bless than or equal to 2

    • one year ago
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