Limit and convergence/divergence of an integral I was working on a problem concerning the function $$f(x) = \frac{x^2}{\ln(x)^\sqrt{x}}$$ asking for the value of $$\lim_{x \to \infty}f(x)$$ and for the convergence/divergence of $$\int_2^\infty f(x) \ dx $$
I wasn't sure if my approach for the limit was entirely valid (or maybe too complicated): $$\lim_{x \to \infty} f(x) = e^{\lim_{x \to \infty} \ln(f(x))} = e^{\lim_{x \to \infty} 2\ln(x) - \sqrt x \ln(\ln(x))}$$ $$\sqrt x > \ln(x)$$ Hence: $$\lim_{x \to \infty} \ln(f(x)) = -\infty \ ; \lim_{x \to \infty} f(x) = e^{-\infty} = 0 \ .$$
As for the integral, I started out by taking the "sum comparison approach" and tried out the root and ratio tests, but those came out inconclusive. I suspect I could use the Limit Comparison Test somehow, but I couldn't find a sufficient "g(x)" to compare to.
Could someone validate/correct my solution for the limit and/or point me in the right direction on the integral question?
Thank you kindly.
 A: For the limit, notice that $f (x)$ is bounded below by $y=0$. Next realize that for $x $ sufficiently large we have that $\ln x>x^{1/3}$ hence for large $x $ we have
$$0 <f (x)<g (x)=\frac {x^2}{x^{\sqrt {x}/3}}=x^{2-\sqrt {x}/3}$$
Then notice that for $x>81$ we have that $2-\sqrt {x}/3<-1$ (it equals $-1$ at $x=81$ and is decreasing from there on out). Hence for large $x $ we have $g (x)<1/x $. We conclude that for large $x $ we have
$$0 <f (x)<\frac {1}{x}$$
Thus by the Squeeze Theorem $f (x) $ converges to $0$.
For the integral notice that for large $x $ we have $f (x)<1/x^2$ which converges (to see this, look at the $g (x) $ we used to bound $f (x) $ and notice that the exponent is decreasing and it reaches $-2$).
A: Let $f(x)=\frac{x^2}{\log^{\sqrt{x}}(x)}$.  Then, we have
$$\begin{align}
\int_2^\infty f(x)\,dx&=\int_2^\infty \frac{x^2}{\log^{\sqrt{x}}(x)}\,dx
\end{align}$$
Enforcing the substitution $x\to x^2$ yields
$$\begin{align}
\int_2^\infty \frac{x^2}{\log^{\sqrt{x}}(x)}\,dx&=2\int_\sqrt{2}^\infty \frac{x^5}{\log^x(x^2)}\,dx\\\\
&=2\int_\sqrt{2}^\infty \frac{x^5}{2^x\log^x(x)}\,dx\\\\
&\le \frac2{\log^\sqrt 2(\sqrt 2)}\int_0^\infty x^5e^{-\log(2)x}\,dx\\\\
&= \frac{240}{\log^\sqrt 2(\sqrt 2)\log^6(2)}\\\\
&<\infty
\end{align}$$
