Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

De limiet $https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 0} \: \frac {x+\sin 2x + \tan 3x} {x - \tan 2x}$ is gelijk aan	A. 1
B. − 3
C. 3
D. − 6
E. 6

The limit $https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 0} \: \frac {x+\sin 2x + \tan 3x} {x - \tan 2x}$ is equal to	A. 1
	B. − 3
	C. 3
	D. − 6
	E. 6

1^ste manier : zonder de regel van DE L' HOSPITAL
$\\\displaystyle\lim_{x\to 0}\:\frac{x+\sin 2x+\tan 3x}{x-\tan 2x}=\displaystyle\lim_{x\to 0}\:\frac{\frac{x}{6x}+\frac{\sin2x}{3.2x}+\frac{\tan3x}{2.3x}}{\frac{x}{6x}-\frac{\tan2x}{3.2x}}\\=\frac{\frac16+\frac13+\frac12}{\frac16-\frac13}=\frac{\frac{1+2+3}{6}}{\frac{1-2}{6}}=\frac{1}{-\frac16}=-6$
2^de manier : met de regel van DE L' HOSPITAL
$\\\displaystyle\lim_{x\to 0}\:\frac{x+\sin 2x+\tan 3x}{x-\tan 2x}\left(=\frac00\right)\buildrel H\over=\displaystyle\lim_{x\to 0}\:\frac{1+2\cos2x+\frac{3}{\cos^23x}}{1-\frac{2}{\cos^2x}}\\=\frac{1+2+3}{1-2}=\frac{6}{-1}=-6$
3^de manier : "officieus" → in bijna alle gevallen kan je sin2x door 2x,
tan 3x door 3x en tan2x door 2x vervangen (als x → 0)
$\\\displaystyle\lim_{x\to 0}\:\frac{x+\sin 2x+\tan 3x}{x-\tan 2x}=\displaystyle\lim_{x\to 0}\:\frac{x+2x+3x}{x-2x}=\displaystyle\lim_{x\to 0}\:\frac{6x}{-x}=-6$