Question

lim ln(tan x) as tends to pi/2 from the left

Answer 1

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

$\displaystyle\lim_{x\to\frac\pi2^-}\ln(\tan x)=\lim_{y\to\infty}\ln y=\infty$