What's the limit as x-> infinity?

• January 9th 2011, 01:44 AM
Boysilver
What's the limit as x-> infinity?
Prove that $x^{1/\ln\ln{x}} \rightarrow \infty$ as $x \rightarrow \infty$.
• January 9th 2011, 01:49 AM
FernandoRevilla
Quote:

Originally Posted by Boysilver
Prove that $x^{1/\ln\ln{x}} \rightarrow \infty$ as $x \rightarrow \infty$.

Using the definition of limit ?.

Fernando Revilla
• January 9th 2011, 01:55 AM
DrSteve
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.
• January 9th 2011, 02:01 AM
alexmahone
Quote:

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'.
• January 9th 2011, 08:23 PM
Drexel28
Quote:

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"