Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

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

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 0} \: \frac {\sin^2x\,+\,1\,-\,\cos x} {x^2}$ is equal to	A. 1
B. 2
C.
D. ∞
E. 0

1^ste manier : zonder de regel van de l'Hospital
$\displaystyle\lim_{x\to 0}\:\frac{\sin^2x\,+\,1\,-\,\cos x}{x^2}=\displaystyle\lim_{x\to 0}\:\frac{\sin^2x}{x^2}+\displaystyle\lim_{x\to 0}\frac{1-\cos x}{x^2}\:\\=\displaystyle\lim_{x\to 0}\:\frac{\sin x}{x}.\displaystyle\lim_{x\to 0}\:\frac{\sin x}{x}+\displaystyle\lim_{x\to 0}\:\frac{(1-\cos x)(1+\cos x)}{x^2(1+\cos x)}=1.1+\displaystyle\lim_{x\to 0}\:\frac{1-\cos^2x}{x^2(1+\cos x)}\\=1+\displaystyle\lim_{x\to 0}\:\frac{\sin^2x}{x^2(1+\cos x)}=1+\displaystyle\lim_{x\to 0}\:\frac{\sin x}{x}.\displaystyle\lim_{x\to 0}\:\frac{\sin x}{x}.\displaystyle\lim_{x\to 0}\:\frac{1}{1+\cos x}\\=1+1.1.\frac{1}{1+1}=1+\frac12=\frac32$
2^de manier : met de regel van de l'Hospital
$\\\displaystyle\lim_{x\to 0}\:\frac{\sin^2x\,+\,1\,-\,\cos x}{x^2}\left(=\frac00\right)\buildrel H\over=\displaystyle\lim_{x\to 0}\:\frac{2\sin x.\cos x+\sin x}{2x}\\\displaystyle\lim_{x\to 0}\:\frac{\sin2x+\sin x}{2x}=\displaystyle\lim_{x\to 0}\:\frac{\sin2x}{x}+\displaystyle\lim_{x\to 0}\:\frac{\sin x}{x}=1+\frac12=\frac32$