Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to \frac{\pi}{4}}\;\frac{1-\tan x}{\sin x-\cos x}=$	A. 1
	B.
	C.
	D. −
	E. −

1^ste manier : zonder de regel van de l'Hospital
$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to \frac{\pi}{4}}\: \frac{1-\tan x}{\sin x - \cos x}=\lim_{x \to \frac{\pi}{4}}\: \frac{\cos x.(1-\tan x)}{\cos x.(\sin x -\cos x)}=\\\lim_{x \to \frac{\pi}{4}}\: \frac{\cos x - \sin x}{\cos x.(\sin x - \cos x)}=-\lim_{x \to \frac{\pi}{4}}\:\frac{1}{\cos x}=-\frac{2}{\sqrt 2}=-\sqrt 2$
2^de manier : met de regel van de l'Hospital
$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to \frac{\pi}{4}}\: \frac{1-\tan x}{\sin x - \cos x}= \lim_{x \to \frac{\pi}{4}}\:\frac{-\frac{1}{\cos^2 x}}{\cos x + \sin x} = \frac{-\frac{1}{\frac 24}}{\frac{\sqrt 2}{2}+\frac{\sqrt 2}{2}} =\frac{-2}{\sqrt 2}=-\sqrt 2$