vertical asymptote of function with $\ln(f(x))$ in the denominator

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How I can find the vertical asymptotic of functions like:

$$y = \frac{\ln^2(x) - 1 }{\ln(x)}$$

$y = \dfrac{2}{\ln(x)}.$ (I didn't understood why the function doesn't have a vertical asymptotic at $x = 0.$)

And what is the technical way and steps needed to find the vertical asymptotic of such functions(without placing numbers to see if the limit tends to infinity...)?

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We need to look at the limit at the points at which $\ln x=0$, that is

$$\lim_{x\to 1^+}\frac{\ln^2 x - 1 }{\ln x}=-\infty$$

$$\lim_{x\to 1-}\frac{\ln^2 x - 1 }{\ln x}=\infty$$

and also

$$\lim_{x\to 0^+}\frac{\ln^2 x - 1 }{\ln x}=\lim_{x\to 0^+} \ln x -\frac1{\ln x}\to -\infty$$

but for you second example

$$\lim_{x\to 0^+}\frac{2}{\ln x}=0$$