Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\large \dpi{110}\displaystyle \lim_{x \to 0}\: \frac{1-\cos x}{x^2}$ is gelijk aan	A. 1
B. 2
C. 0
D.
E.

$https://latex.codecogs.com/png.image?\large \dpi{110}\displaystyle \lim_{x \to 0}\: \frac{1-\cos x}{x^2}$ is equal to	A. 1
	B. 2
	C. 0
	D. ∞
	E.

1^ste manier : zonder de regel van de l'Hospital
$\\\displaystyle\lim_{x\to 0}\:\frac{1-\cos x}{x^2}=\displaystyle\lim_{x\to 0}\:\frac{(1-\cos x)(1+\cos x)}{x^2(1+\cos x)}=\displaystyle\lim_{x\to 0}\:\frac{1-\cos^2x}{x^2(1+\cos x)}=\displaystyle\lim_{x\to 0}\:\frac{\sin^2x}{x^2(1+\cos x)}\\=\displaystyle\lim_{x\to 0}\:\frac{\sin x}{x}.\displaystyle\lim_{x\to 0}\:\frac{\sin x}{x}.\displaystyle\lim_{x\to 0}\:\frac{1}{1+\cos x}=1.1.\frac{1}{1+1}=\frac12$
2^de manier : met de regel van de l'Hospital
$\\\displaystyle\lim_{x\to 0}\:\frac{1-\cos x}{x^2}=\left(=\frac00\right)\overset{H}{=}\displaystyle\lim_{x\to 0}\:\frac{0+\sin x}{2x}=\frac12.\displaystyle\lim_{x\to 0}\:\frac{\sin x}{x}=\frac12$