Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to \frac{\pi}{2}} \: \frac {\sin 2x} {\cos 3x}$ is gelijk aan	A. 1
B. 2
C.
D. −
E.

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to \frac{\pi}{2}} \: \frac {\sin 2x} {\cos 3x}$ is equal to	A. 1
	B. 2
	C.
	D. −
	E.

1^ste manier : met de formule voor cos 3x, nl. cos 3x = 4cos³ − 3cos x
$\lim_{x\to\frac{\pi}{2}}\:\frac{\sin 2x}{\cos 3x}=\lim_{x\to\frac{\pi}{2}}\:\frac{2\sin x\cos x}{4\cos^3x-3\cos x}=\lim_{x\to\frac{\pi}{2}}\:\frac{2\sin x}{4\cos^2x-3}=\frac{2.1}{4.0^2-3}=-\frac23$
2^de manier :
$\lim_{x\to\frac{\pi}{2}}\:\frac{\sin 2x}{\cos 3x}\left(=\frac 00\right)\overset{H}{=}\lim_{x\to\frac{\pi}{2}}\:\frac{2\cos 2x}{-3\sin 3x}=\frac{2\cos\pi}{-3\sin\frac32\pi}=\frac{2.(-1)}{-3.(-1)}=)\frac23$