# Limit of sequence n^p/e^n as n approaches infinity

1. Oct 20, 2011

### alex_dc1

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. Oct 20, 2011

### alex_dc1

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?

Last edited: Oct 20, 2011
3. Oct 20, 2011

### lineintegral1

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.

4. Oct 20, 2011

### alex_dc1

I've never heard that term before. Would that basically state that the p-th derivative of n^p equals (p!)?

5. Oct 21, 2011

### HallsofIvy

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.