Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to \pi}\: \frac{\sin x - \sin \pi}{x-\pi}$ is gelijk aan	A. − 1
B. 0
C. 1
D. π
E. −π

$\displaystyle \lim_{x \to \pi}\: \frac{\sin x - \sin \pi}{x-\pi}$ >is equal to	A. − 1
	B. 0
	C. 1
	D. π
	E. −π

1^ste manier : zonder de regel van de l'Hospital
$\\\displaystyle\lim_{x\to\pi}\;\frac{\sin x-\sin\pi}{x-\pi}\overset{S}{=}\;\displaystyle\lim_{x\to\pi}\;\frac{2\cos\frac{x+\pi}{2}.\sin\frac{x-\pi}{2}}{x-\pi}\:=\displaystyle\lim_{x\to\pi}\;\cos\frac{x+\pi}{2}.\displaystyle\lim_{x\to\pi}\;\frac{\sin\frac{x-\pi}{2}}{\frac{x-\pi}{2}}\\=\;\cos\frac{\pi+\pi}{2}.1=-1$
2^de manier : met de regel van de l'Hospital
$\\\displaystyle\lim_{x\to\pi}\;\frac{\sin x-\sin\pi}{x-\pi}\left(=\frac00\right)\overset{H}{=}\;\displaystyle\lim_{x\to\pi}\;\frac{\cos x-0}{1-0}=cos\pi=-1$