Can anybody think of a straightforward way to see that this limit equals $0$ for all $A > 1$? I can only think of applying L'Hopital's rule, which leads to a godawful mess. Perhaps there's a useful upper bound for $n^{\lg n}$?
Limit of $n^{A\lg n}\cdot 2^{n-n^A}$ for $A > 1$
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limits
logarithms
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0Limit... as what? $n\to\infty$? – 2017-01-01
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0Is $\lg$ supposed to be the base 2 log? Or did you mean to use $\ln$, the natural log? – 2017-01-01
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0Maybe use $n^{Alog(n)} = (2^{log(n)})^{A log(n)}$ (I'm assuming the log is in base 2, but this is applicable for any base). – 2017-01-01
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0Yes, as n approaches infinity. And yes, I do mean the base 2 log. – 2017-01-01
1 Answers
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$$n^{A\lg n}=2^{A\ln^2n}$$
Thus, we may rewrite it as
$$2^{A\lg^2n+n-n^A}$$
As $n\to\infty$ the $-n^A$ dominates, since $A\lg^2n=\mathcal O(n)$ and $n^A>n^1$. Thus, as $n\to\infty$, the limit is simply $0$.