How do I evaluate the limit $\large\lim_{x\to \infty }\frac{\ln(x)^{\ln(x)^{\ln(x)}}}{x^x}$? $$\large\lim_{x\to \infty }\frac{\ln(x)^{\ln(x)^{\ln(x)}}}{x^x}$$
As $x$ approaches infinity, both functions approach infinity. Therefore I should use the hopital rule, right? But it seems to complicate the answer.
 A: Take the $\ln$ of the bottom. We get $x\ln x$. Do it again. We get $\ln x+\ln\ln x$.
Take the $\ln$ of the top. We get $(\ln x)^{\ln x}\ln\ln x$. Do it again. We get $\ln x\ln\ln x+\ln\ln\ln x$.
The ratio of the $\ln\ln$'s is
$$\frac{\ln x\ln\ln x+\ln\ln\ln x}{\ln x+\ln\ln x}.\tag{1}$$
This $\to\infty$  as $x\to\infty$. To see that, divide top and bottom of (1) by $\ln x$. So the original ration goes to infinity. 
A: Going the opposite way compared to André Nicolas:
The basic idea is that $a^b = e^{b \ln a}$.
Do this once with $a = \ln x$, $b = \ln(x)^{\ln(x)}$ :
$\ln(x)^{\ln(x)^{\ln(x)}}
=e^{\ln(x)^{\ln(x)}\ln \ln (x)}
$.
Do this again with $a = \ln x$, $b = \ln(x)^{\ln(x)}$ :
$\ln(x)^{\ln(x)}
=e^{\ln(x)\ln \ln (x)}
$,
so
$$\ln(x)^{\ln(x)^{\ln(x)}}
=e^{\ln(x)^{\ln(x)}\ln \ln (x)}
=e^{e^{\ln(x)\ln \ln (x)}\ln \ln (x)}
=e^{e^{\ln(x)\ln \ln (x)+\ln \ln \ln (x)}}
$$
Now look at $x^x$.
$$x^x = e^{x \ln(x)}
=e^{e^{\ln(x)}\ln(x)}
=e^{e^{\ln(x)+\ln\ln(x)}}
$$
The second level exponents
are
$\ln(x)\ln \ln (x)+\ln \ln \ln (x)$
and 
$\ln(x)+\ln\ln(x)$
and the first is clearly larger
since
$\dfrac{\ln(x)\ln \ln (x)+\ln \ln \ln (x)}{\ln(x)+\ln\ln(x)}
\approx \ln \ln (x) \to \infty$
as $x \to \infty$.
Looking at these two solutions,
I think André Nicolas's is better
because it is easier to understand.
A: Let $\ln x=y\implies x=e^y$
then limit converts to $$\lim_{y\to\infty}\frac{y^{y^y}}{e^{ye^y}}$$
Now compare powers of $y,e$ in numerator and denominator as $y\uparrow$
$\lim_{y\to\infty}\frac{y^y}{ye^y}=\lim_{y\to\infty}\frac{y^{y-1}}{e^y}\rightarrow\infty$
Now in original limit $$\lim_{y\to\infty}\frac{y^{y^y}}{e^{ye^y}}$$
$y$ is surely $> e$ as $y\uparrow \infty$ and power of $y$ in numerator also grows much faster than power of $e$ in denominator
so $$\lim_{y\to\infty}\frac{y^{y^y}}{e^{ye^y}}>\lim_{y\to\infty}\frac{e^{y^y}}{e^{ye^y}}\to\infty$$
