# Putting z=x-pi/4 show that limit of (1-tanx)/(1-sqrt2 sinx)=2 as x approaches to pi/4?

Dec 12, 2017

$x = z + \frac{\pi}{4}$

$\tan x = \tan \left(z + \frac{\pi}{4}\right)$

$= \frac{\tan z + \tan \frac{\pi}{4}}{1 - \tan z \tan \frac{\pi}{4}}$

$= \frac{\tan z + 1}{1 - \tan z \left(1\right)}$

$= \frac{1 + \tan z}{1 - \tan z}$

$\sin x = \sin \left(z + \frac{\pi}{4}\right)$

$= \sin z \cos \frac{\pi}{4} + \cos z \sin \frac{\pi}{4}$

$= \frac{1}{\sqrt{2}} \left(\sin z + \cos z\right)$

${\lim}_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{1 - \sqrt{2} \sin x}$

$= {\lim}_{z \to 0} \frac{1 - \frac{1 + \tan z}{1 - \tan z}}{1 - \sqrt{2} \frac{1}{\sqrt{2}} \left(\sin z + \cos z\right)}$

$= {\lim}_{z \to 0} \frac{\frac{1 - \tan z - \left(1 + \tan z\right)}{1 - \tan z}}{1 - \sin z - \cos z}$

$= {\lim}_{z \to 0} \frac{- 2 \tan z}{\left(1 - \sin z - \cos z\right) \left(1 - \tan z\right)}$

(Assuming limit exists)

$= {\lim}_{z \to 0} \frac{- 2 \tan z}{1 - \sin z - \cos z}$

$= 2 {\lim}_{z \to 0} \frac{\tan z}{\sin z + \cos z - 1}$

$= 2 {\lim}_{z \to 0} {\left(\frac{\sin z + \cos z - 1}{\tan} z\right)}^{- 1}$

$= 2 {\lim}_{z \to 0} {\left(\sin \frac{z}{\tan} z - \frac{1 - \cos z}{\tan} z\right)}^{- 1}$

$= 2 {\lim}_{z \to 0} {\left(\cos z - \frac{1 - {\cos}^{2} z}{\tan z \left(1 + \cos z\right)}\right)}^{- 1}$

$= 2 {\lim}_{z \to 0} {\left(1 - \frac{{\sin}^{2} z}{\tan z \left(1 + \cos z\right)}\right)}^{- 1}$

$= 2 {\lim}_{z \to 0} {\left(1 - \frac{\sin z \cos z}{1 + \cos z}\right)}^{- 1}$

$= 2 {\lim}_{z \to 0} {\left(1 - \frac{\left(0\right) \left(1\right)}{1 + 1}\right)}^{- 1}$

$= 2$