Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

chrisplusian

  • one year ago

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

  • This Question is Closed
  1. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    What is \(p\)?

  2. chrisplusian
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    P>0 I really dont even understand why they give that information accept that it shows n remains in the numerator

  3. chrisplusian
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I tried to use logarithmic properties to find an answer but it just kept running me in circles

  4. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[n^p \times \frac{1}{e^{n}}\]\(e\) 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.

  5. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Am I right? O_O

  6. terenzreignz
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Definitely not the reasoning I'd pick...

  7. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Knew it.

  8. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    correct answer but wrong logic ... think about the opposite e^n/n^p and n->infinity

  9. terenzreignz
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I prefer simple cases... \[\Large 2e^n \rightarrow \infty\]\[\Large \frac1{e^n}\rightarrow 0\] \[\Large 2e^n \times \frac1{e^n }\rightarrow 0 \quad \color{red}?\]

  10. chrisplusian
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  11. DLS
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    0

  12. chrisplusian
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Sorry (1/e^n)

  13. terenzreignz
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Relax... you can either use L'Hôpital or one of my favourite theorems~ pick one

  14. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Approaching infinity, not really infinity. I tried to use l'hopital: didn't work. :-|

  15. terenzreignz
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The almighty Squeeze~

  16. DLS
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Isnt the answer 0?

  17. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ARMAGERD. How could I forget the squeeze

  18. chrisplusian
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ok I tried lopitals and it did not work nor did logarithmic properties

  19. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    What's the inequality BTW?

  20. chrisplusian
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @DLS the answer is zero

  21. DLS
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Yes,its easy.

  22. DLS
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Use LH

  23. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @DLS how?

  24. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    My favouraite technique is \[ a = e^{\log a} \\ n^{p} = e^{p \log n } \]

  25. terenzreignz
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  26. terenzreignz
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    q is guaranteed to be a positive integer, means it [the numerator] will (eventually) differentiate to zero...

  27. chrisplusian
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @experimentX why would that help?

  28. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    L'hopital works use it least floor(p + 1)

  29. terenzreignz
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Why deviate from the norm, @experimentX ? :D Just use ceiling :3

  30. DLS
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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: \[y=5x^4\] \[\frac{dy}{dx}=20x^3=>60x^2=>120x=>120\]

  31. chrisplusian
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Don't understand what you mean @experimentX

  32. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @chrisplusian |dw:1369325101292:dw|

  33. terenzreignz
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  34. terenzreignz
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    And then squeezing :)

  35. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @terenzreignz yes ceil

  36. DLS
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    @chrisplusian my reply should probably clear it off. @experimentX does it seem right?

  37. chrisplusian
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the solutions guide says =0 when p>0, and bless than or equal to 2

  38. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.