improper integrals - Convergence of $int_1^2 frac{1}{sqrt ln x} mathrm { d}x$ by Asymptotic Comparison Test.

According to the material I have,
If $f:[a,b)\to \mathbb{R} ,b\in \mathbb{R},f\geq0 $
Then:
i) If $f$ is an infinite function of real order $\alpha\lt1$ with respect to $\frac {1}{x-b}$ $\implies$ $\int_a^bf(x)\mathrm { d}x\lt\infty$
ii) If $f$ is an infinite function of real order $\alpha\geq1$ with respect to $\frac {1}{x-b}$ $\implies$ $\int_a^bf(x)\mathrm { d}x=\infty$
Based on this information I am trying to evaluate the convergence of $\int_1^2 \frac{1}{\sqrt \ln x} \mathrm { d}x$
In order to determine the order of my function I consider $$\lim_{x\to1+}{\frac{1}{\sqrt\ln x \over\frac{1}{(x-2)^\alpha}}}=\lim_{x\to1+}\frac {(x-2)^\alpha}{\sqrt\ln x}$$
Despite trying various methods I haven't been able to obtain a real number as a limit value for some $\alpha\in\mathbb{R}$

Answer

An issue is that you look at $b=2$. You should consider the fact that the "problem" is at $1$, and adapt your theorem (it's easy) for $f\colon (a,b]\to\mathbb{R}$ and look at the $a=1$ end.

Rewrite, for $x>1$,
$$
\sqrt{\ln x} = \sqrt{\ln(1+(x-1))}\,.
$$
This seems silly, but now, recall that

$$
\lim_{u\to 0} \frac{\ln(1+u)}{u} = 1
$$
so that
$$
\lim_{x\to 1} \frac{\sqrt{\ln x}}{(x-1)^{1/2}} = \lim_{x\to 1} \sqrt{\frac{\ln x}{x-1}}= \lim_{x\to 1} \sqrt{\frac{\ln(1+(x-1))}{x-1}} = \sqrt{\lim_{x\to 1} \frac{\ln(1+(x-1))}{x-1}} = \sqrt{1}=1
$$

Can you conclude?