Respuesta :

Let [tex]x=\arctan y[/tex]. Then [tex]\tan x=y[/tex], and so as [tex]x\to\dfrac\pi2^-[/tex], you have [tex]y\to+\infty[/tex]. The limit is then equivalent to

[tex]\displaystyle\lim_{x\to\frac\pi2^-}\ln(\tan x)=\lim_{y\to\infty}\ln y=\infty[/tex]