Help solving a limit as n approaches infinity?

Question

Take the limit as n approaches infinity of (n^p / e^n) where p>0


The answer in my workbook is 0 
where p>0 and  n (greater than or equal to) 2

But I have no idea how to get to that.


Steps or an explanation, please and thank you.

kitty123 · Accepted Answer

n^p is a polynomial and e^n is an exponential. An exponential grows much much faster then a polynomial. Hence, as n approaches infinity, e^n will approach infinity faster than n^p. If the denominator goes to infinity first then (n^p/e^n) would approach zero.

I think this is also one of the standard limits.

If the question had been e^n/n^p then the limit have would been undefined.

Geeganage W · Answer

Let n = 1/x; when n→∞, x→0

n→∞, (n^p / e^n) = n→0  n^p/(1 +n + n^2/2! +n^3/3! + ...)
                         = n→0 1/(1/n^p +1/n^(p-1)2!+1/n^(p-2)3! +...+...)
                         = 1/ ∞ =0

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Help solving a limit as n approaches infinity?

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