# what is $\displaystyle \lim_{n\to \infty }\ \left(\frac{1}{n}\right)^\frac{1}{\text{ln}\text{ ln}(n)}$

how can I find the next limit please?

I tried some sentences but they don't help me.

The font is so small, so the exercise is: (1/n)^(1/ln(ln(n))).

$$\lim_{n\to \infty } \left(\frac{1}{n}\right)^\frac{1}{\ln\ln(n)}$$

Maybe let $u$ to be: $\ln(n)$ and then $u' = 1/n$...? I really don't know..

Thank you all!

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Taking logarithms gives $$\log \left(\frac 1n\right)^{1/\log\log n} = \frac 1{\log\log n} \cdot \log\frac 1n = - \frac{\log n}{\log \log n}$$ Now as $\frac{m}{\log m} \to \infty$ for $m \to\infty$ and $\log n \to \infty$ for $n \to \infty$, we have

$$\log \left(\frac 1n\right)^{1/\log\log n} = - \frac{\log n}{\log \log n} \to -\infty, \qquad n \to \infty$$ hence $$\left(\frac 1n\right)^{1/\log\log n} \to 0, \qquad n \to \infty$$

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wow, thank you! although I didn't learn the operating of 'log', now I know it thanks to you! thank you! –  Alon Shmiel Nov 23 '12 at 15:17

Let $u_n= \log \log n$. Then $u_n \to \infty$ and $n= e^{e^{u_n}}$.

Then

$$\lim_{n\to \infty} (\frac{1}{n})^{\frac{1}{\ln\ln(n)}}= \lim_n \frac{1}{e^{\frac{e^{u_n}}{u_n}}}$$

Now use the fact that $\lim_{x \to \infty} \frac{e^x}{x}= \infty$.

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