What's the limit as x-> infinity?

**Boysilver** · January 9th 2011, 12:44 AM

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

**FernandoRevilla** · January 9th 2011, 12:49 AM

Originally Posted by Boysilver

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

Using the definition of limit ?.

Fernando Revilla

**DrSteve** · January 9th 2011, 12:55 AM

I'm guessing that you want to show this using L'Hopital's rule?

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

Take the $\ln$ of both sides.

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

Now see if you can finish it with L'Hopital's Rule.

**alexmahone** · January 9th 2011, 01:01 AM

Originally Posted by Boysilver

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

I have a 'handwaving' solution.

Let $x=10^{100}$

$ln x=230$ (approx.)

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

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

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

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

PS: This may not fetch you any marks in an exam but it will help you to 'think in limits'.

**Drexel28** · January 9th 2011, 07:23 PM

Originally Posted by Boysilver

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

So, we are trying to ascertain $\displaystyle \lim_{x\to\infty} x^{\frac{1}{\log(\log(x))}$ . Let $\log(\log(x))=y$ so that our limit becomes $\displaystyle \lim_{y\to\infty}(e^{e^{y})^{\frac{1}{y}}$ and thus to assume our limit existed would be to assume that $\displaystyle \lim_{y\to\infty}\frac{e^y}{y}$ existed. More formally, $x^{\frac{1}{\log(\log(x))}}=e^{\frac{\log(x)}{\log (\log(x))}}\succ e^{\sqrt{\log(x)}}\to \infty$ where $\succ$ means "asymptotically dominates"

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