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Wolfram tells me that the the limit is $0$ when $n$ goes to infinity. Unfortunately, I have no idea how to prove it...

$$\lim_{n\to\infty}\frac {2^\sqrt { \log(\log n)}}{\log n}.$$

Any help would be appreciated, thanks in advance.

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Try finding the limit of the logarithm of that expression. – Gerry Myerson Oct 26 '12 at 11:59
try l'hopital's rule – user31280 Oct 26 '12 at 12:02
@F'OlaYinka Or not. – Did Oct 26 '12 at 12:42
First, realize that you can solve the simpler: $\frac{2^\sqrt{\log m}}{m}$ by setting $m=\log n$ – Thomas Andrews Oct 26 '12 at 12:53
up vote 2 down vote accepted


  1. The logarithm of this quantity is $\log 2\cdot\sqrt{\log(\log n)}-\log(\log n)$.

  2. When $n\to+\infty$, $\log(\log n)\longrightarrow$ $_________$.

  3. When $x\to+\infty$, $\log2\cdot\sqrt{x}-x\longrightarrow$ $_________$.

  4. Hence $\log2\cdot\sqrt{\log(\log n)}-\log(\log n)\longrightarrow$ $________$ when $n\to+\infty$.

  5. And finally $2^{\sqrt{\log(\log n)}}/\log n\longrightarrow$ $_________$ when $n\to+\infty$.

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Starting with $$f(n) = \frac{2^{\sqrt{\log(\log(n))}}}{\log(n)}$$ and taking logarithms $$\log(f(n)) = \log{2}{\sqrt{\log(\log(n))}} - \log({\log(n)})=\sqrt{\log(\log(n))}\left(\log(2)-\sqrt{\log(\log(n))}\right)$$

then $\sqrt{\log(\log(n))}$ increases towards $+\infty$ with increasing $n$, while $\left(\log(2)-\sqrt{\log(\log(n))}\right)$ heads towards $-\infty$, so $\log(f(n))$ heads towards $-\infty$ and $f(n)$ heads towards $0$.

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