Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{{x \to 0}}\; \frac {x\,\cdot\,\sin 2x} {1\,-\,\cos 3x}$ is gelijk aan	A. \(\frac89\)
B. \(\frac23\)
C. \(\frac13\)
D. \(\frac49\)
E. \(1\)

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{{x \to 0}}\; \frac {x\,\cdot\,\sin 2x} {1\,-\,\cos 3x}$ is equal to	A. $\frac89$
	B. $\frac23$
	C. $\frac13$
	D. $\frac49$
	E. $1$

$Daar\;\;1-\cos 2x=2.\sin^2x\;\;is\;\;1-\cos 3x=2.sin^2\frac{3x}{2}$
We houden ook rekening met
$\displaystyle\lim_{x\to0}\frac{ax}{\sin bx}=\lim_{x\to0}\frac{\sin ax}{\sin bx}=\frac ab$
1^ste manier : zonder de regel van de l'Hospital
$\\\displaystyle\lim_{{x\to 0}}\;\frac{x\,\cdot\,\sin 2x}{1\,-\,\cos 3x}=\lim_{{x\to 0}}\;\frac{x.\sin2x}{2\sin^2\frac{3x}{2}}=\lim_{{x\to 0}}\;\frac{2x\sin x\cos x}{2\sin^2\frac{3x}{2}}\\=\lim_{{x\to 0}}\;\frac{x}{\sin\frac{3x}{2}}.\lim_{{x\to 0}}\;\frac{\sin x}{\sin\frac{3x}{2}}.\lim_{{x\to 0}}\;\cos x=\frac23.\frac23.1=\frac49$
2^de manier : met de regel van de l'Hospital
$\\\displaystyle\lim_{{x\to 0}}\;\frac{x\,\cdot\,\sin 2x}{1\,-\,\cos 3x}\left(=\frac00\right)\buildrel H\over=\lim_{{x\to 0}}\;\frac{\sin2x+x.(\cos 2x).2}{3.\sin3x}\\=\lim_{{x\to 0}}\;\frac{\sin2x}{3\sin3x}+\lim_{{x\to 0}}\;\frac{2x}{3\sin3x}.\lim_{{x\to 0}}\;\cos2x=\frac13.\frac23+\frac13.\frac23.1=\frac29+\frac29=\frac49$