Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to \frac{\pi }{2}}\:\frac {\cos x} {\pi\:-\:2x}$ is gelijk aan	A. 1
B.
C.
D. − 1
E.

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to \frac{\pi }{2}}\:\frac {\cos x} {\pi\:-\:2x}$ is equal to	A. 1
	B.
	C.
	D. − 1
	E.

1^ste manier : zonder de regel van de l'Hospital
Stel y = x − pi/2

, dan zal y → 0
$\\\displaystyle\lim_{x\to\frac{\pi}{2}}\:\frac{\cos x}{\pi\:-\:2x}=\displaystyle\lim_{x\to\frac{\pi}{2}}\:\frac{\cos\left(y+\frac{\pi}{2}\right)}{\pi-2\left(y+\frac{\pi}{2}\right)}=\displaystyle\lim_{x\to\frac{\pi}{2}}\:\frac{-\sin y}{\pi-2y-\pi}\\=\displaystyle\lim_{x\to\frac{\pi}{2}}\:\frac{-\sin y}{-2y}=\frac12.\displaystyle\lim_{x\to\frac{\pi}{2}}\:\frac{\sin y}{y}=\frac12$
2^de manier : met de regel van de l'Hospital
$\\\displaystyle\lim_{x\to\frac{\pi}{2}}\:\frac{\cos x}{\pi\:-\:2x}\left(=\frac00\right)\overset{H}{=}\displaystyle\lim_{x\to\frac{\pi}{2}}\:\frac{-\sin x}{-2}=\frac{-1}{-2}=\frac12$