Wiskunde Multiple Choice Vraag

De limiet $\displaystyle\lim_{x\to\,3}\frac{\frac{1}{\sqrt3}-\frac{1}{\sqrt x}}{x-3}$ is gelijk aan	A. \(\frac{1}{3}\)
B. \(6\sqrt3\)
C. \(3\)
D. \(\frac{\sqrt3}{18}\)
E. \(\frac{\sqrt3}{6}\)

1^ste manier : (zonder de regel van de l’Hospital)
$\inline\\\displaystyle\lim_{x\to\,3}\frac{\frac{1}{\sqrt3}-\frac{1}{\sqrt x}}{x-3}=\lim_{x\to\,3}\frac{\frac{\sqrt x-\sqrt3}{\sqrt{3x}}}{x-3}=\lim_{x\to\,3}\frac{\sqrt x-\sqrt3}{\sqrt{3x}.(\sqrt x-\sqrt3)(\sqrt x+\sqrt3)}\\=\lim_{x\to\,3}\frac{1}{\sqrt{3x}.(\sqrt x+\sqrt3)}=\frac{1}{3(\sqrt3+\sqrt3)}=\frac{1}{6\sqrt3}=\frac{\sqrt3}{18}$
2^de manier : (met de regel van de l’Hospital)
$\displaystyle\lim_{x\to\,3}\frac{\frac{1}{\sqrt3}-\frac{1}{\sqrt x}}{x-3}\left(=\frac{0}{0}\right)\overset{H}{=}\lim_{x\to\,3}\frac{0+\frac{1}{x}.\frac{1}{2\sqrt x}}{1-0}=\frac{1}{6\sqrt3}=\frac{\sqrt3}{18}$