Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$\displaystyle \lim_{x \to 0}\frac{1-\cos x + x.\sin x}{2x^2}=$	A. 1
B.
C.
D.
E.

1^ste manier : zonder de regel van de l'Hospital
$\\\displaystyle \lim_{x \to 0}\frac{1-\cos x + x.\sin x}{2x^2}=\displaystyle \lim_{x \to 0}\frac{2.\sin^2 \frac x2+ x.\sin x}{2x^2}=\\\displaystyle \lim_{x \to 0}\frac{\sin^2 \frac x2} {x^2}+\displaystyle \lim_{x \to 0}\frac{\sin x} {2x}=\frac 14 + \frac 12 = \frac 34$
2^de manier : met de regel van de l'Hospital
$\\\displaystyle \lim_{x \to 0}\frac{1-\cos x + x.\sin x}{2x^2}\left ( =\frac 00 \right )\overset{H}{=}\displaystyle \lim_{x \to 0}\frac{\sin x+ \sin x + x.\cos x }{4x}\\=\displaystyle \lim_{x \to 0}\frac{2\sin x}{4x}+\displaystyle \lim_{x \to 0} \frac{\cos x}{4}=\frac 24 + \frac 14 = \frac 34$