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$\displaystyle\lim_{n\to\infty} \frac{2^\sqrt{\log n}}{n\log^3 n}$

At first, I tried L'Hopital's Rule, but that appeared to make things more complicated. Then I tried this:

$\displaystyle\lim_{n\to\infty} e^{\log2\log n - \log(\log^3 n)}$

But I don't see how to proceed from there.

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Put $log(n) = x$. Then as $n \rightarrow \infty, x \rightarrow \infty$.

Our new limit is $\lim_{x \to \infty} \frac {2^{\sqrt(x)}}{e^x x^3}$, which gives an indeterminate form.

Let's once again use a substitution. Put $u = \sqrt{x}$. Then as $x \to \infty, u \to \infty$. We rewrite the limit as $\lim_{u \to \infty} \frac{2^u}{e^{u^2}u^{6}}$. Then this becomes $\lim_{u \to \infty} \frac{1}{u^6} * \lim_{u \to \infty} ({\frac {2}{e^u})}^u$ = $\lim_{u \to \infty} \frac{1}{u^6} * \lim_{u \to \infty} ({\frac {2}{e^u})}^u$.

Let us deal now with $\lim_{u \to \infty} ({\frac {2}{e^u})}^u$.

Let $y = \lim_{u \to \infty} ({\frac {2}{e^u})}^u$. Then $log(y) = \lim_{u \to \infty} u log(\frac {2}{e^u}) = \lim_{u \to \infty} u \{log 2 - log (e^u)\} = \lim_{u \to \infty}\{ulog2 - u^2\},$ which diverges to $-\infty$. But this tell us $y \rightarrow 0$, so that we now have $\lim_{u \to \infty} \frac{1}{u^6} * 0 = 0*0 = 0,$ as desired.

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In order to prove that the limit is zero, it is enough to show that

(1) For any $n$ large enough, $2^{\sqrt{\log n}}<n$.

The inequality is equivalent to $\sqrt{\log(n)} \log(2) < \log(n)$, or to $\log(2)<\sqrt{\log(n)}$, or to $$ n > \exp\left(\log^2(2)\right). $$

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Use the substitution $n\mapsto e^{n^2}$: $$ \begin{align} \lim_{n\to\infty}\frac{2^\sqrt{\log(n)}}{n\log^3(n)} &=\lim_{n\to\infty}\frac{2^n}{e^{n^2}n^6}\\ &=\lim_{n\to\infty}e^{n\log(2)-n^2-6\log(n)}\\ &=0 \end{align} $$ Since $$ \lim_{n\to\infty}n\log(2)-n^2-6\log(n)=-\infty $$

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