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Homework Help: Limit of sequence n^p/e^n as n approaches infinity

  1. Oct 20, 2011 #1
    Simple question just as the title says, but I can't remember or derive the solution for the life of me. I know that the answer is 0. I know why the answer is 0. But I need to know the mathematical derivation of the solution, and that's the part that I can't remember. So, to reiterate, how do you find:

    lim (n^p)/(e^n) as n→∞, where p>0
  2. jcsd
  3. Oct 20, 2011 #2
    Wait, I think I just figured it out, but I want to make sure this makes sense:

    (n^p)/(e^n) =

    n/e^(n/p) * n/e^(n/p) * n/e^(n/p) * n/e^(n/p).......


    lim n/e^(n/p) as n→∞ is (by L'hopital's rule) equal to:

    *edit for the correct derivative:
    lim 1/[ (1/p) * e^(n/p) ] n →∞ is equal to 0, therefore:

    n/e^(n/p) * n/e^(n/p) * n/e^(n/p) * n/e^(n/p)....... is equal to 0 * 0 * 0 * 0..... therefore,

    lim (n^p)/(e^n) as n→∞ is equal to 0

    Does that make sense, or am I just making crap up?
    Last edited: Oct 20, 2011
  4. Oct 20, 2011 #3
    Um, p-iterations of L'Hospital's rule would be much clearer, I think. The limit is certainly 0, but I think you are making it more complicated than it needs to be.
  5. Oct 20, 2011 #4
    I've never heard that term before. Would that basically state that the p-th derivative of n^p equals (p!)?
  6. Oct 21, 2011 #5


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    That reminds me of the "proof" that the derivative of x^2 is just "x". Since x^n is just x added to itself n times, x^x is x added to itself x times: x^x= (x+ x+ x+ ...+ x) and therefore the derivative is (x^x)'= (1+ 1+ 1+ ...+ 1) x times which is equal to x.

    That's wrong because you cannot treat doing an operation a variable number of times the same as doing it a constant number of times.

    lineintegral is suggesting that you use L'Hopital's rule p times (that's what "p iteration" means). Each time you differentiate the numerator you get a lower power of nwhile the denominator is always e^n. After differentiating p times you have just p! in the numerator and e^n in the denominator. Now take the limit as n goes to infinity.

    Strictly speaking, that only works for p a positive integer. For the more general case, write n^p as e^(ln(n^p)= e^(p ln(n)) so that n^p/e^n= e^(pln(n)- n) and look at the limit of pln(n)- n as n goes to infinity.
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