Wiskunde Multiple Choice Vraag

$\displaystyle\lim_{x\to\,0}\:\frac{x^3+\sin^3x}{x^3}=$	A. 1
B. 2
C. 3
D. 0
E.

1^ste manier : de snelste en eenvoudigste manier
$\displaystyle\lim_{x\to\,0}\:\frac{x^3+\sin^3x}{x^3}=\lim_{x\to\,0}\:\left(1+\frac{\sin^3 x}{x^3}\right)=1+1^3=2$
2^de manier : met de regel van de l'Hospital
$\inline\\\displaystyle\lim_{x\to\,0}\:\frac{x^3+\sin^3x}{x^3}\left(=\frac00\right)\overset{H}{=}\lim_{x\to\,0}\:\frac{3x^2+3\sin^2x.\cos x}{3x^2}\\=1+\lim_{x\to\,0}\:\frac{\sin^2x.\cos x}{x^2}=1+1^2.1=2$
3^de manier : zonder de regel van De L'Hospital
$\inline\\\displaystyle\lim_{x\to\,0}\:\frac{x^3+\sin^3x}{x^3}=\lim_{x\to\,0}\:\frac{x+\sin x}{x}\cdot\lim_{x\to\,0}\:\frac{x^2-x.\sin x+\sin^2x}{x^2}\\=(1+1)\cdot\lim_{x\to\,0}\:\left(1-\frac{\sin x}{x}+\frac{\sin^2x}{x^2}\right)\\=2\cdot(1-1+1^2)=2\cdot 1=2$

^{gricha - vA153 - 30.3.2026}