take the limit as N approaches infinity 2^(n)/n^(2) I know when you take the lim
ID: 3118141 • Letter: T
Question
take the limit as N approaches infinity 2^(n)/n^(2)
I know when you take the limit of this you get infinity/infinity then you use Lhopital's rule to take the derivative I think I am messing that up somehow.
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Explanation / Answer
lim of (2^n)/(n^2) = infinity / infinity as n -> inf differentiate both the numerator and denom. with respect to n : let y = 2^n => lny = n ln2 => 1/y y' = ln2 => y' = y ln2 = 2^n ln2 and d/dn (n^2) = 2n so the limit becomes : lim [(2^n *ln(2)] / 2n = inf /inf n -> inf so differentiate again both num. and den. : d/dn [(2^n *ln(2)] = [ln(2)]^2 * 2^n and d/dn (2n) = 2 so the limit becomes lim [(2^n *(ln(2))^2] / 2 n -> inf = infinity
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