Wiskunde Multiple Choice Vraag

$\displaystyle \lim_{x\to3}\frac{\sqrt{x+6}-3}{x^3-27}=$	A. \(1\)
B. \(\frac19\)
C. \(\frac1{81}\)
D. \(\frac1{162}\)
E. \(\frac1{216}\)

1^ste manier : zonder de regel van de L'Hospital
$\inline\\ \displaystyle \lim_{x\to3}\frac{\sqrt{x+6}-3}{x^3-27}=\lim_{x\to3}\frac{(\sqrt{x+6}-3)(\sqrt{x+6}+3)}{(x^3-27)(\sqrt{x+6}+3)}\\=\lim_{x\to3}\frac{x+6-9}{(x-3)(x^2+3x+9)(\sqrt{x+6}+3)}\\=\frac{1}{(9+9+9)(\sqrt{9}+3)}=\frac{1}{27.6}=\frac{1}{162}$
2^de manier : met de regel van de L'Hospital
$\displaystyle \lim_{x\to3}\frac{\sqrt{x+6}-3}{x^3-27}\left(=\frac00\right)\overset{H}{=}\lim_{x\to3}\frac{\frac{1}{2\sqrt{x+6}}}{3x^2}=\frac{\frac{1}{2.3}}{27}=\frac{1}{6.27}=\frac{1}{162}$