Wiskunde Multiple Choice Vraag

$\displaystyle\lim_{x\to0}\;\frac{2-\cos x-\cos^2x}{x^2}=$	A. −1
B. 0
C. 1
D. 1,5
E. 2,5

1^ste manier : zonder de regel van de L'Hospital
$\inline\\\displaystyle\lim_{x\to0}\;\frac{2-\cos x-\cos^2x}{x^2}=\lim_{x\to0}\;\frac{1-\cos x+1-\cos^2x}{x^2}\\=\lim_{x\to0}\;\frac{1-\cos x}{x^2}+\lim_{x\to0}\;\frac{1-\cos^2x}{x^2}\\=\lim_{x\to0}\;\frac{(1-\cos x)(1+\cos x)}{x^2\cdot(1+\cos x)}+\lim_{x\to0}\;\frac{\sin^2x}{x^2}\\=\lim_{x\to0}\;\frac{1-\cos^2x}{x^2\cdot(1+\cos x)}+\lim_{x\to0}\;\left(\frac{\sin x}{x}\right)^2\\=\lim_{x\to0}\;\frac{\sin^2x}{x^2\cdot(1+\cos x)}+\left(\lim_{x\to0}\;\frac{\sin x}{x}\right)^2\\=\left(\lim_{x\to0}\;\frac{\sin x}{x}\right)^2\cdot\;\lim_{x\to0}\;\frac{1}{1+\cos x}+\left(\lim_{x\to0}\;\frac{\sin x}{x}\right)^2\\=1^2\cdot\frac{1}{1+1}+1^2=1\cdot\frac12+1=\frac32$
2^de manier : met de regel van de L'Hospital
$\inline\\\displaystyle\lim_{x\to0}\;\frac{2-\cos x-\cos^2x}{x^2}\left(=\frac00\right)\overset{H}{=}\lim_{x\to0}\;\frac{\sin x+2\cos x.\sin x}{2x}\\=\lim_{x\to0}\;\frac{\sin x}{2x}+\lim_{x\to0}\;\cos x\cdot\lim_{x\to0}\;\frac{\sin x}{x}=\frac12+1.1=\frac32$

^{gricha - vA162 - 30.3.2026}