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

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

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

• @Daman I don't get it. What should I write?
– EL02
Commented Jan 4, 2021 at 14:48

3 Answers

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

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

• +1 nice and clean Commented Jan 4, 2021 at 14:57
• Why $1-\tan x=\frac{y}{1+y}$
– EL02
Commented Jan 4, 2021 at 15:12
• Try finding $\tan x$ first. Commented Jan 4, 2021 at 15:35
• I'm agree with @gt6989b Commented Jan 4, 2021 at 20:40

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?

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