Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 0}\: \frac{\cot 2x}{\cot x}=$	A. = 1
B. = 2
C. =
D. bestaat niet
E. = 0,5

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 0}\: \frac{\cot 2x}{\cot x}=$	A. = 1
B. = 2
C. = ∞
D. does not exist
E. = 0.5

$\text{We houden rekening met}\;\displaystyle\lim_{x\to 0}\;\frac{\sin ax}{\sin bx}=\lim_{x\to 0}\;\frac{\tan ax}{\tan bx}=\frac ab$
1^ste manier : zonder de regel van de l'Hospital
$\\\lim_{x\to 0}\:\frac{\cot 2x}{\cot x}=\lim_{x\to 0}\:\frac{\frac{\cos 2x}{\sin 2x}}{\frac{\cos x}{\sin x}}=\lim_{x\to 0}\:\frac{\cos 2x.\sin x}{\sin 2x.\cos x}=\lim_{x\to 0}\:\frac{\cos 2x.\sin x}{2.\cos^2x.\sin x}=\frac{1}{2.1^2}=\frac12$
2^de manier : zonder de regel van de l'Hospital
$\\\lim_{x\to 0}\:\frac{\cot 2x}{\cot x}=\lim_{x\to 0}\:\frac{\tan x}{\tan 2x}=\frac12$
3^de manier : met de regel van de l'Hospital
$\\\lim_{x\to 0}\:\frac{\cot 2x}{\cot x}\left(=\frac{\pm\infty}{\pm\infty}\right)\overset{H}{=}\lim_{x\to 0}\:\frac{-\frac{1}{\sin^22x}.2}{-\frac{1}{\sin^2x}}=\lim_{x\to 0}\:\frac{2\sin^2x}{\sin^22x}=2.\lim_{x\to 0}\:\left(\frac{\sin x}{\sin2x}.\frac{\sin x}{\sin 2x}\right)\\=2.\frac12.\frac12=\frac12$