Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 0} \: \frac {1\,-\,\cos 4x} {\tan^2\, 3x}$ is gelijk aan	A.
B.
C.
D.
E. geen van de vorige

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 0} \: \frac {1\,-\,\cos 4x} {\tan^2\, 3x}$ is equal to	A. ∞
B.
C.
D.
E. none of the above

Bij de drie manieren houden we rekening met $\displaystyle\lim_{x\to0}\frac{\sin ax}{\tan bx}=\frac ab $
1^ste manier :
$\displaystyle\lim_{x\to 0}\:\frac{1\,-\,\cos 4x}{\tan^2\,3x}=\displaystyle\lim_{x\to 0}\:\frac{(1-cos4x)(1+cos4x)}{\tan^23x(1+\cos4x)}=\displaystyle\lim_{x\to 0}\:\frac{1-\cos^24x}{\tan^23x.(1+\cos4x)}\\=\displaystyle\lim_{x\to 0}\,\frac{\sin^24x}{\tan^23x.(1\!+\!\cos4x)}=\displaystyle\lim_{x\to 0}\,\frac{1}{1\!+\!\cos4x}.\left(\displaystyle\lim_{x\to 0}\,\frac{\sin4x}{\tan3x}\right)^2\!\!=\frac{1}{1+1}.\frac43.\frac43=\frac89$
2^de manier : rekening houdend de formule 1 − cos 2x = 2sin² x
$\displaystyle\lim_{x\to 0}\:\frac{1\,-\,\cos 4x}{\tan^2\,3x}=\displaystyle\lim_{x\to 0}\:\frac{2\sin^22x}{\tan^23x}=2\displaystyle\lim_{x\to 0}\:\frac{\sin2x}{\tan3x}.\displaystyle\lim_{x\to 0}\:\frac{\sin2x}{\tan3x}=2.\frac23.\frac23=\frac89$
3^de manier : met de regel van de l'Hospital
$\\\displaystyle\lim_{x\to 0}\:\frac{1\,-\,\cos 4x}{\tan^2\,3x}\left(=\frac00\right)\buildrel H\over=\displaystyle\lim_{x\to 0}\:\frac{4\sin4x}{2\tan3x\frac{1}{\cos^23x}.3}\\=\frac46\displaystyle\lim_{x\to 0}\:\frac{\sin4x.\cos^33x}{\tan3x}=\frac23.\frac43.1=\frac89$