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Lim of An=n^2*exp(-sqrt(n))

by oferon
Tags: ann2expsqrtn
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oferon
#1
Feb11-12, 03:28 PM
P: 30
Hi all, my problem regards this limit:

[tex]\lim_{n\to\infty}n^2e^{(-\sqrt{n})}[/tex]

Obviously equals 0, but I can't find how to show it.
Tried the squeeze theorem (coudn't find any propriate upper bound)
Ratio test won't seem to work..
I do realize the reason for that is that the set approaches 0 starting at heigher n's..

Anyway.. how can I prove convergence and find the limit in a formal way? thanks!
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mathman
#2
Feb11-12, 04:12 PM
Sci Advisor
P: 6,040
Simple method: Let m=√n, so the problem is limit m -> ∞ m4/em.

em = 1 + m + m2/2! + m3/3! + m4/4! + m5/5! + .... It is obvious from the 5th term on the denominator of the fraction swamps the numerator.
oferon
#3
Feb11-12, 05:56 PM
P: 30
I've tried changing variables like you did and got m4/em, which does seem nicer..
But is using taylor expansion the only way to solve here?
I'm pretty sure that's not what the course staff expected us to do..

mathman
#4
Feb12-12, 05:53 PM
Sci Advisor
P: 6,040
Lim of An=n^2*exp(-sqrt(n))

Have learned L'Hopital's rule?
If so, use that. Take 5 derivatives of the numerator and the denominator and get 0/em.


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