What's the $\lim _{x \to \frac{π}{4}} \frac{\ln (\cot x)}{{1-\tan x}}$?

Question

$$ \lim_{x \to \pi/4} \frac{\ln (\cot x)}{{1-\tan x}} $$

I'm trying to solve it without using L'Hopital or Taylor series.

Answer 1

Set $1 + y = \cot x$. Then $1 - \tan x = \frac y{1+y}$, and hence

$$\lim_{x \to \pi/4} \frac {\ln \cot x}{1 - \tan x} = \lim_{y \to 0} \frac {\ln (1+y)}{y}(1+y) = 1\cdot 1 = 1$$

Answer 2

\begin{gather*} \varliminf _{x\rightarrow \frac{\pi }{4}}\frac{\ln( \cot x)}{1-\tan x} =\varliminf _{h\rightarrow 0}\frac{\ln\left( \cot\left(\frac{\pi }{4} -h\right)\right)}{1-\tan\left(\frac{\pi }{4} -h\right)}\\ =\varliminf _{h\rightarrow 0}\frac{\ln\left(\frac{1+\tan( h)}{1-\tan( h)}\right)}{1-\frac{1-\tan( h)}{1+\tan( h)}} =\varliminf _{h\rightarrow 0}\frac{\ln\left( 1+\frac{2\tan( h)}{1-\tan( h)}\right)}{\frac{2\tan( h)}{1+\tan( h)}}\\ \end{gather*} Can you take it from here?

Answer 3

Let the given limit be $L$ and $y = \tan x$. Then observe that as $x \to \pi /4 $ we have $y \to 1$. Then $$ L = \lim_{x \to \pi /4} \frac{ \ln \cot x } {1 - \tan x} = \lim_{y \to 1} \frac{ \ln y}{y - 1} $$

Can you finish this?