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Lim of An=n^2*exp(sqrt(n))by oferon
Tags: ann2expsqrtn 
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#1
Feb1112, 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! 


#2
Feb1112, 04:12 PM

Sci Advisor
P: 6,040

Simple method: Let m=√n, so the problem is limit m > ∞ m^{4}/e^{m}.
e^{m} = 1 + m + m^{2}/2! + m^{3}/3! + m^{4}/4! + m^{5}/5! + .... It is obvious from the 5th term on the denominator of the fraction swamps the numerator. 


#3
Feb1112, 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.. 


#4
Feb1212, 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/e^{m}. 


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