Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$\lim_{x \to 0 }\; \frac {(x-2)^3+8} {4x}$ is gelijk aan	A. 1
B. 2
C. 3
D. 4
E. 8

$\lim_{x \to 0 }\; \frac {(x-2)^3+8} {4x}$ is equal to	A. 1
B. 2
C. 3
D. 4
E. 8

1^ste manier :
$\\\lim_{x\to 0}\;\frac{(x-2)^3+8}{4x}=\lim_{x\to 0}\;\frac{x^3-3x^2.2+3x.2^2-8+8}{4x}=\lim_{x\to 0}\;\frac{x^3-6x^2+12x}{4x}\\=\lim_{x\to 0}\;\frac{x^2-6x\!+12}{4x}=\frac{12}{4}=3$
2^de manier :
$\\\lim_{x\to 0}\;\frac{(x-2)^3+8}{4x}=\lim_{x\to 0}\;\frac{(x-2+2)[(x-2)^2-2(x-2)+2^2]}{4x}=\frac{(0-2)^2-3(0-2)+4}{4}\\=\frac{4+4+4}{4}=3$
3^de manier :
$\\\lim_{x\to 0}\;\frac{(x-2)^3+8}{4x}\left(=\frac00\right)\overset{H}{=}\lim_{x\to 0}\;\frac{3(x-2)^2}{4}=\frac{3.(-2)^2}{4}=3$