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Could somebody check if this is correct?

$$\lim_{n \to \infty} \frac {\log_{2}(\log_{2}(n))}{\log_{2}(n)}$$

I exponantiate the numerator and the denominator with 2

$$\frac {(\log_{2}(\log_{2}(n)))^2}{(\log_{2}(n))^2}$$

$$ = \frac {\log_{2}(n)}{n}$$

I extract the constant from the logarithm

$$ = \log_{2}(e) * \lim_{n \to \infty} \frac {\ln(n)}{n}$$

Using de l'Hospital:

$$ = \log_{2}(e) * \lim_{n \to \infty} \frac {\frac {1}{n}}{1}$$

$$ = \log_{2}(e) * \lim_{n \to \infty} \frac {1}{n} = 0$$

Is that correct?

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That is not what squaring does to logarithms, and either of the things you could be trying to do does not preserve limits. Try a substitution like m = log n instead. –  Qiaochu Yuan Jan 9 '11 at 23:52
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3 Answers

up vote 3 down vote accepted

Instead of squaring, it's possible to dive right into L'Hôpital's rule, since as $n\to\infty$, you get the indeterminate form $\infty/\infty$, as you seem to have noticed. It doesn't make much sense to 'extract the constant from the logarithm,' but instead you should have $$ \lim_{n\to\infty}\frac{\log_2(\log_2(n))}{\log_2(n)}=\lim_{n\to\infty}\frac{\log_2(\log_2(n))'}{\log_2(n)'}=\lim_{n\to\infty}\frac{\frac{1}{n\ln{2}\ln{n}}}{\frac{1}{n\ln{2}}}=\lim_{n\to\infty}\frac{1}{\ln{n}}. $$

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OMG this is so obvious. Thanks! –  learning_sam Jan 10 '11 at 0:33
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HINT $\ $ Let $\rm\ n = 2^{2^k}\:.\ $ Then the limit becomes that of $\rm\ k/2^k\ $ as $\rm\ k\to \infty$

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Let $ t = \log_{2}(n) $ then $ \lim \limits_{n \to \infty} \frac {\log_{2}(\log_{2}(n))}{\log_{2}(n)}=\lim \limits_{t \to \infty} \frac {\log_{2}(t)}{t}=0$

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