Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

De limiet $https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 0} \: \frac {\cos x\:-\:1} {x}$ bestaat (?) en is gelijk aan	A. 1
B. 2
C.
D. 0
E. bestaat niet

IN CONSTRUCTION	A.
B.
C.
D.
E.

$\\\displaystyle\lim_{x\to 0}\:\frac{\cos x\:-\:1}{x}=\displaystyle\lim_{x\to 0}\:\frac{(\cos x\:-\:1)(\cos x+1)}{x.(\cos x+1)}=\displaystyle\lim_{x\to 0}\:\frac{\cos^2x-1}{x.(\cos x+1)}\\=\displaystyle\lim_{x\to 0}\:\frac{-\sin^2 x}{x.(\cos x+1)}=-\displaystyle\lim_{x\to 0}\:\frac{\sin x}{x}.\displaystyle\lim_{x\to 0}\:\frac{\sin x}{\cos x+1}=-1.\frac02=0$

^{gricha - v6675 - 10.5.15}