Wiskunde Multiple Choice Vraag

$\displaystyle\lim_{x\to 0}\:\frac{x-\sin x}{x\cdot\sin^2x}=$	A. \(1\)
B. \(\frac12\)
C. \(\frac13\)
D. \(\frac14\)
E. \(\frac16\)

$\inline\displaystyle\lim_{x\to 0}\:\frac{x-\sin x}{x\cdot\sin^2x}\left(=\frac00\right)\overset{H}{=}\lim_{x\to 0}\:\frac{1-\cos x}{\sin^2x+x(2\sin x\cos x)}=\lim_{x\to 0}\:\frac{1-\cos x}{\sin^2x+x.\sin2x}\\\overset{H}{=}\lim_{x\to 0}\:\frac{\sin x}{2\sin x.\cos x+\sin 2x+2x\cos2x}\overset{H}{=}\lim_{x\to 0}\:\frac{1}{2\cos x+2\cos x+2\frac{x}{\sin x}.\cos x}\\=\frac{1}{2.1+2.1+2.1.1}=\frac16$