Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 4} \: {\frac{x^2-16}{2-\sqrt x}}$ is gelijk aan	A. − 32
B. − 24
C. − 16
D. 0
E. 8
F. 32

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 4} \: {\frac{x^2-16}{2-\sqrt x}}$ is equal to	A. − 32
	B. − 24
	C. − 16
	D. 0
	E. 8
	F. 32

1^ste manier :
$\\\displaystyle\lim_{x\to 4}\:{\frac{x^2-16}{2-\sqrt x}}=\displaystyle\lim_{x\to 4}\:\frac{(x-4)(x+4)}{2-\sqrt x}=\displaystyle\lim_{x\to 4}\:\frac{(\sqrt x-2)(\sqrt x+2)(x+4)}{2-\sqrt x}\\=\displaystyle\lim_{x\to 4}\:(\sqrt x+2)(x+4)=-(2+2)(4+4)=-4.8=-32$
2^de manier :
$\\\displaystyle\lim_{x\to 4}\:{\frac{x^2-16}{2-\sqrt x}}=\displaystyle\lim_{x\to 4}\:\frac{(x^2-16)(2+\sqrt x)}{(2-\sqrt x)(2+\sqrt x)}=\displaystyle\lim_{x\to 4}\:\frac{(x-4)(x+4)(2+\sqrt x)}{4-x}\\=\displaystyle\lim_{x\to 4}\:(x+4)(2+\sqrt x)=-(4+4)(2+2)=-8.4=-32$
3^de manier :
$\\\displaystyle\lim_{x\to 4}\:{\frac{x^2-16}{2-\sqrt x}}\left(=\frac00\right)\overset{H}{=}\displaystyle\lim_{x\to 4}\:\frac{2x-0}{0-\frac{1}{2\sqrt x}}=\displaystyle\lim_{x\to 4}\:(-4x\sqrt x)=-4.4.2=-32$