Wiskunde Multiple Choice Vraag

$\displaystyle\lim_{x\to0}\:\frac{2x-\sin2x}{3x-\sin3x}=$	A. \(1\)
B. \(\frac23\)
C. \(\frac49\)
D. \(\frac8{27}\)
E. \(\frac12\)

$\inline\\\displaystyle\lim_{x\to\,0}\frac{2x-\sin2x}{3x-\sin3x}\left(=\frac00\right)\overset{H}{=}\lim_{x\to\,0}\frac{2-2\cos2x}{3-3\cos3x}\left(=\frac00\right)\overset{H}{=}\lim_{x\to\,0}\frac{0+4\sin2x}{0+9\sin3x}\\=\frac49\lim_{x\to\,0}\frac{\sin2x}{\sin3x}=\frac49\lim_{x\to\,0}\frac{2x\frac{\sin2x}{2x}}{3x\frac{\sin3x}{3x}}=\frac49.\frac23.\frac{\displaystyle\lim_{x\to\,0}\frac{\sin2x}{2x}}{\displaystyle\lim_{x\to\,0}\frac{\sin3x}{3x}}=\frac{8}{27}.\frac11=\frac{8}{27}=0,\overline{296}$