Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 0}\frac {1\,-\,\cos 2x} {1\:-\:\frac{\sin x}{x}}$ is gelijk aan	A. 1
B. 2
C. 4
D. 8
E. 12

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 0}\frac {1\,-\,\cos 2x} {1\:-\:\frac{\sin x}{x}}$ is equal to	A. 1
B. 2
C. 4
D. 8
E. 12

1^ste manier : met de regel van de L'Hospital
$\\\lim_{x\to 0}\frac{1\,-\,\cos 2x}{1\:-\:\frac{\sin x}{x}}=\lim_{x\to 0}\frac{x-x\cos2x}{x-\sin x}\left(=\frac00\right)\\\buildrel H\over=\lim_{x\to 0}\frac{1-\cos2x+2x\sin2x}{1-\cos x}\left(=\frac00\right)\\\buildrel H\over=\lim_{x\to 0}\frac{0+2\sin2x+2\sin2x+4x\cos2x}{\sin x}\\=2\lim_{x\to 0}\frac{\sin2x}{\sin x}+2\lim_{x\to 0}\frac{\sin2x}{\sin x}+4\lim_{x\to 0}\frac{x}{\sin x}.\cos 2x\\=2.2+2.2+4.1.1=4+4+4=12$
2^de manier : ook met de regel van de L'Hospital
$\\\lim_{x\to 0}\frac{1\,-\,\cos 2x}{1\:-\:\frac{\sin x}{x}}=\lim_{x\to 0}\frac{2\sin^2x}{1-\frac{\sin x}{x}}=\lim_{x\to 0}\frac{2x\sin^2x}{x-\sin x}\left(=\frac00\right)\\\buildrel H\over=\frac{2\sin^2x+2x.2.\sin x.\cos x}{1-\cos x}\\=2\lim_{x\to 0}\frac{\sin^2x+x\sin2x}{1-\cos x}\left(=\frac00\right)\\\buildrel H\over=2\lim_{x\to 0}\frac{2\sin x.\cos x+\sin2x+2x.\cos2x}{\sin x}\\=4\lim_{x\to 0}\frac{\sin x.\cos x+\sin x.\cos x+x.\cos2x}{\sin x}\\=4\lim_{x\to 0}\left(\cos x+\cos x+\frac{x}{\sin x}.\cos2x\right)\\=4(1+1+1+1)=4.3=12$