Solve for $\lim_{x \rightarrow \infty} \frac{x^{\ln(x)}}{(\ln(x))^x}$

Question

Solve for
$$ \lim_{x \rightarrow \infty} \frac{x^{\ln(x)}}{(\ln(x))^x} $$

I used l'Hôpital's rule but in vain. Could you help me someway. Thanks for attention.

Answer 1

hint: prove $\dfrac{x^{\ln x}}{(\ln x)^x} < \dfrac{1}{x}$ and this implies the limit is $0$. To this end, we prove: $x^{1+\ln x} < (\ln x)^x\iff \ln x(1+\ln x) < x\ln(\ln x)$. This final inequality can be established by taking derivative of the difference of LHS and RHS, and is left for you to try out. I think it should work.

Answer 2

Let $f(x)=\frac{ x^{\ln(x)}}{ (\ln(x))^x }$ be your function.

If we take logarithm, we get

$$g(x)=\ln(f(x))$$

$$=(\ln(x))^2-x\ln(\ln(x))$$

$$=x\left(\frac{(\ln(x))^2}{x}-\ln(\ln(x)\right)$$

when $x\to +\infty$, the power grows fastly than logarithm.

$$\implies \lim_{x\to +\infty}\frac{(\ln(x))^2}{x}=0$$

$$\implies \lim_{x\to +\infty}g(x)=-\infty$$

$$\implies \lim_{x\to +\infty}f(x)=0.$$

Answer 3

Hint:

Using continuity of exponential and logarithm we have: $$ \lim_{x\to + \infty}\frac{x^{\ln x}}{\ln^x x}= \exp\left[\ln\left(\lim_{x\to + \infty}\frac{x^{\ln x}}{\ln^x x} \right) \right]= $$ $$ =\exp\left[\lim_{x\to + \infty}\left(\ln^2 x-x \ln(\ln x) \right) \right] $$ Now it is not difficult to show that the limit in the square brakets is $-\infty$