Wiskunde Multiple Choice Vraag

$\displaystyle\lim_{x\to 0}\frac{\tan^2 2x}{1-\cos 2x}=$	A. 1
B. 2
C. 0
D. 4
E. 0,5

$\displaystyle\lim_{x\to 0}\frac{\tan^22x}{1-\cos2x}\left(=\frac00\right)\overset{H}{=}\displaystyle\lim_{x\to 0}\frac{2\tan 2x.\frac{1}{\cos^22x}.2}{0+2\sin 2x}=\displaystyle\lim_{x\to 0}\frac{4.\frac{\sin 2x}{\cos^32x}}{2\sin 2x}=2.\displaystyle\lim_{x\to 0}\frac{1}{\cos^32x}=2.1=2$