Hello Xcessive,
We are given to compute:
$$L=\lim_{x\to\infty}\frac{x^{\ln(x)}}{\ln^x(x)}$$
Observing that we have the indeterminate form $$\frac{\infty}{\infty}$$, we may apply L'Hôpital's Rule. Thus, we need to compute the derivatives of the numerator and denominator.
Let's begin with the numerator, and let:
$$y=x^{\ln(x)}$$
Taking the natural log of both sides, we obtain:
$$\ln(y)=\ln^2(x)$$
Implicitly differentiating with respect to $x$, we have:
$$\frac{1}{y}\frac{dy}{dx}=\frac{2}{x}\ln(x)$$
$$\frac{dy}{dx}=\frac{2y}{x}\ln(x)$$
$$\frac{dy}{dx}=\frac{2x^{\ln(x)}}{x}\ln(x)=2x^{\ln(x)-1}\ln(x)$$
Now, for the denominator, let:
$$y=\ln^x(x)$$
Taking the natural log of both sides, we have:
$$\ln(y)=x\ln(\ln(x))$$
Implicitly differentiating with respect to $x$, we have:
$$\frac{1}{y}\frac{dy}{dx}=x\frac{1}{\ln(x)}\frac{1}{x}+\ln(\ln(x))=\ln(\ln(x))+\frac{1}{\ln(x)}$$
$$\frac{dy}{dx}=y\left(\ln(\ln(x))+\frac{1}{\ln(x)} \right)$$
$$\frac{dy}{dx}=\ln^x(x)\left(\ln(\ln(x))+\frac{1}{\ln(x)} \right)$$
And so we may now state:
$$L=\lim_{x\to\infty}\frac{2x^{\ln(x)-1}\ln(x)}{\ln^x(x) \left(\ln(\ln(x))+\frac{1}{\ln(x)} \right)}= 2L\lim_{x\to\infty}\frac{\ln(x)}{x\left(\ln(\ln(x))+\frac{1}{\ln(x)} \right)}$$
At this point, we may observe that the denominator dominates the numerator, and so we may conclude:
$$L=0$$