How do I evaluate the limit $\large\lim_{x\to \infty }\frac{\ln(x)^{\ln(x)^{\ln(x)}}}{x^x}$?

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.

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.

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 $$\lim_{y\to\infty}\frac{y^{y^y}}{e^{ye^y}}>\lim_{y\to\infty}\frac{e^{y^y}}{e^{ye^y}}\to\infty$$