Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$\displaystyle \lim_{x \to 0} \: \frac {\sin^2 x} {1-\,\cos x}=$	A. 1
	B. 2
	C. − 1
	D. ∞
	E. 0.5

1^ste manier :
$\\\lim_{x\to 0}\:\frac{\sin^2 x}{1-\,\cos x}=\lim_{x\to 0}\:\frac{1-\cos^2x}{1-\cos x}=\lim_{x\to 0}\:\frac{(1-\cos x)(1+\cos x)}{1-\cos x}\\\\=\lim_{x\to 0}\:(1+\cos x)=1+1=2$
2^de manier :
$\\\lim_{x\to 0}\:\frac{\sin^2 x}{1-\,\cos x}\left(=\frac00\right)\overset{H}{=}\lim_{x\to 0}\:\frac{2.\sin x.\cos x}{\sin x}=\lim_{x\to 0}\:2\cos x=2.1=2$