How to calculate $ \lim_{x\to\infty}\frac{x^{\ln x}}{(\ln x)^x} $ How to calculate 
$$
\lim_{x\to\infty}\frac{x^{\ln x}}{(\ln x)^x}
$$
Using L'Hopital rule comes out too complicated. Thanks!  
 A: L'Hospital's rule does not solve all limits (and should be avoided, generally speaking: when it works, it is equivalent to Taylor's formula at order $1$, and most students forget to  check validity hypotheses for L'Hospital).
Hint:
Rewrite the fraction as 
$$\frac{\mathrm e^{\ln^2x}}{\mathrm e^{x\ln\ln x}}=\mathrm e^{\ln^2x-x\ln\ln x}=\mathrm e^{-x\ln\ln x\big(1-\tfrac{\ln^2x}{x\ln\ln x}\big)}$$
and show $\;\ln^2x=o\bigl(x\ln\ln x\bigr)$.
A: Let $x=e^u$.  Then
$${x^{\ln x}\over{(\ln x)^x}}={e^{u^2}\over u^{e^u}}={e^{u^2}\over e^{e^u\ln u}}=e^{u^2-e^u\ln u}$$
Is it clear that $u^2-e^u\ln u\to-\infty$ as $u\to\infty$?  If not, then it's easy enough to show that $u^2\lt e^u$ for large $u$, hence $u^2-e^u\ln u\lt e^u(1-\ln u)$.
A: It often helps to take the log of an expression, especially one with complicated exponents. Let $f(x)=x^{\log x}/(\log x)^x$ for $x>0.$
For $x>1$ we have $\log x>0$ and $\log f(x)=(\log x)^2-x\log \log x.$ 
For $x>1$ let $x=e^y$ with $y>0,$ so $\log f(x)=y^2-e^y\log y$ which $\to -\infty$ as $y\to \infty$ because for $y>e$ we have $y^2-e^y\log y<y^2-e^y.$
Since $y=\log x \to \infty$ as $x\to \infty,$ therefore $\log f(x)\to -\infty$ as $x\to \infty.$ Therefore $f(x)\to 0$ as $x\to \infty.$
P.S. To show that $y^2-e^y\to -\infty$ as $y\to \infty$: Let $g(y)=e^y.$ For $y>0 $ there exists $z\in (0,y)$ with  $g(y)=g(0)+yg'(0)+\frac {1}{2}y^2g''(0)+\frac {1}{6}y^3g'''(z) =1+y+\frac {1}{2}y^2+\frac {1}{6}y^3e^z>\frac {1}{6}y^3e^z>\frac {1}{6}y^3.$ 
