Question

$\underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\frac{1-\tan x}{x-{\displaystyle \frac{\mathrm{\pi }}{4}}}$

Answer 1

$$\begin{gathered} \underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\frac{1-\tan x}{x-{\displaystyle \frac{\mathrm{\pi }}{4}}} \\ =\underset{h\to 0}{\lim }\frac{1-\tan \left({\displaystyle \frac{\mathrm{\pi }}{4}}+h\right)}{\left({\displaystyle \frac{\mathrm{\pi }}{4}}+h-{\displaystyle \frac{\mathrm{\pi }}{4}}\right)} \\ =\underset{h\to 0}{\lim }\frac{1-\left({\displaystyle \frac{1+\tan h}{1-\tan h}}\right)}{h}\left[\because \tan \left(\frac{\mathrm{\pi }}{4}+0\right)=\frac{1+\tan \theta }{1-\tan \theta }\right] \\ =\underset{h\to 0}{\lim }\frac{1-\tan h-1-\tan h}{\left(1-\tan h\right)h} \\ =\underset{h\to 0}{\lim }\frac{-2\tan h}{h\left(1-\tan h\right)}\left[\because \underset{h\to 0}{\lim }\frac{\tan h}{h}=1\right] \\ \Rightarrow \frac{-2}{1-0}=-2 \end{gathered}$$