Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to \frac{\pi }{2}}\left ( \sin 3x.\cos x.\tan 3x \right )=$	A. −
B.
C. − 3
D. 3
E. − 9
F. 9

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to \frac{\pi }{2}}\left ( \sin 3x.\cos x.\tan 3x \right )=$	A. −
B.
C. − 3
D. 3
E. − 9
F. 9

1^ste manier :
$\\\displaystyle\lim_{x\to\frac{\pi}{2}}\left(\sin 3x.\cos x.\tan 3x\right)=\displaystyle\lim_{x\to\frac{\pi}{2}}\left(\sin^23x.\frac{\cos x}{\cos(2x+x)}\right)\\=\displaystyle\lim_{x\to\frac{\pi}{2}}(\sin3x).\displaystyle\lim_{x\to\frac{\pi}{2}}\left(\frac{\cos x}{\cos 2x.\cos x-\sin 2x.\sin x}\right)\\=\displaystyle\lim_{x\to\frac{\pi}{2}}(\sin^23x).\displaystyle\lim_{x\to\frac{\pi}{2}}\left(\frac{1}{\cos2x-2\sin^2x}\right)\\=(-1)^2.\frac{1}{-1-2.1^2}=-\frac13$
2^de manier : met de formule voor cos 3x
$\\\displaystyle\lim_{x\to\frac{\pi}{2}}\left(\sin 3x.\cos x.\tan 3x\right)=\displaystyle\lim_{x\to\frac{\pi}{2}}\left(\sin^23x.\frac{\cos x}{\cos(3x)}\right)\\=\displaystyle\lim_{x\to\frac{\pi}{2}}(\sin^23x).\displaystyle\lim_{x\to\frac{\pi}{2}}\left(\frac{\cos x}{4\cos^3x-3\cos x}\right)\\=\displaystyle\lim_{x\to\frac{\pi}{2}}(\sin^23x).\displaystyle\lim_{x\to\frac{\pi}{2}}\left(\frac{1}{4\cos^2x-3}\right)=(-1)^2\frac{1}{4.0-3}=-\frac13$
3^de manier : met de regel van de l'Hospital
$\\\displaystyle\lim_{x\to\frac{\pi}{2}}\left(\sin 3x.\cos x.\tan 3x\right)=\displaystyle\lim_{x\to\frac{\pi}{2}}(\sin^23x).\displaystyle\lim_{x\to\frac{\pi}{2}}\left(\frac{\cos x}{\cos 3x}\right)\\\left(=(-1)^2.\frac00\right)\buildrel H\over=1.\displaystyle\lim_{x\to\frac{\pi}{2}}\left(\frac{-\sin x}{-3\sin3x}\right)=\frac{-1}{-3.(-1)}=-\frac13$