Prove that $\displaystyle x^{1/\ln\ln{x}} \rightarrow \infty$ as $\displaystyle x \rightarrow \infty$.

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- Jan 9th 2011, 12:44 AMBoysilverWhat's the limit as x-> infinity?
Prove that $\displaystyle x^{1/\ln\ln{x}} \rightarrow \infty$ as $\displaystyle x \rightarrow \infty$.

- Jan 9th 2011, 12:49 AMFernandoRevilla

Using the definition of limit ?.

Fernando Revilla - Jan 9th 2011, 12:55 AMDrSteve
I'm guessing that you want to show this using L'Hopital's rule?

Let $\displaystyle y=x^{\frac{1}{\ln \ln x}}$

Take the $\displaystyle \ln$ of both sides.

$\displaystyle \ln y=\frac{1}{\ln \ln x}\ln x$.

Now see if you can finish it with L'Hopital's Rule. - Jan 9th 2011, 01:01 AMalexmahone
I have a 'handwaving' solution.

Let $\displaystyle x=10^{100}$

$\displaystyle ln x=230$ (approx.)

$\displaystyle ln ln x=ln 230=5.44$ (approx.)

$\displaystyle \frac{1}{ln ln x}=\frac{1}{5.44}=0.18$ (approx.)

$\displaystyle x^{\frac{1}{ln ln x}}=(10^{100})^{0.18}=10^{18}$

Therefore, $\displaystyle x^{1/\ln\ln{x}} \rightarrow \infty$ as $\displaystyle x \rightarrow \infty$

PS: This may not fetch you any marks in an exam but it will help you to 'think in limits'. - Jan 9th 2011, 07:23 PMDrexel28
So, we are trying to ascertain $\displaystyle \displaystyle \lim_{x\to\infty} x^{\frac{1}{\log(\log(x))}$. Let $\displaystyle \log(\log(x))=y$ so that our limit becomes $\displaystyle \displaystyle \lim_{y\to\infty}(e^{e^{y})^{\frac{1}{y}}$ and thus to assume our limit existed would be to assume that $\displaystyle \displaystyle \lim_{y\to\infty}\frac{e^y}{y}$ existed. More formally, $\displaystyle x^{\frac{1}{\log(\log(x))}}=e^{\frac{\log(x)}{\log (\log(x))}}\succ e^{\sqrt{\log(x)}}\to \infty$ where $\displaystyle \succ$ means "asymptotically dominates"