Wiskunde Multiple Choice Vraag

$\displaystyle\lim_{x\to\,6}\frac {2-\sqrt{x-2}} {x^2-36}=$	A. 1
B.
C. \(-\,\frac{1}{12}\)
D. \(-\,\frac{1}{18}\)
E. \(-\,\frac{1}{48}\)

$\inline\displaystyle\lim_{x\to\,6}\frac{2-\sqrt{x-2}}{x^2-36}\left(=\frac00\right)=\lim_{x\to6}\frac{\left(2-\sqrt{x-2}\right).\left(2+\sqrt{x-2}\right)}{\left(x^2-36\right)\left(2+\sqrt{x-2}\right)}\\=\lim_{x\to6}\frac{4-(x-2)}{(x-6)(x+6)\left(2+\sqrt{x-2}\right)}=\lim_{x\to6}\frac{6-x}{(x-6)(x+6)\left(2+\sqrt{x-2}\right)}\\=\lim_{x\to6}\frac{-1}{(x+6)\left(2+\sqrt{x-2}\right)}=\frac{-1}{12.(2+\sqrt4)}=-\,\frac{1}{48}$