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lim (n^p)/(e^n) as n→∞, where p>0

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lim (n^p)/(e^n) as n→∞, where p>0

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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).......

where:

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?

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

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

where:

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:

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HallsofIvy

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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.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).......

where:

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?

That's wrong because you cannot treat doing an operation a

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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