Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 4} \: \: {\frac{x-4}{\frac{1}{\sqrt x}-\frac 12 }}$ is gelijk aan	A. −16
B. − 8
C. − 4
D. − 2
E. een positief getal

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 4} \: \: {\frac{x-4}{\frac{1}{\sqrt x}-\frac 12 }}=$	A. −16
	B. − 8
	C. − 4
	D. − 2
	E. a positive number

1^ste manier : (zonder de regel van de l'Hospital)
$\displaystyle\lim_{x\to 4}\frac{x-4}{\frac{1}{\sqrt x}-\frac{1}{2}}=\displaystyle\lim_{x\to 4}\frac{x-4}{\frac{2-\sqrt x}{2\sqrt x}}=\displaystyle\lim_{x\to 4}\frac{2\sqrt x(x-4)}{2-\sqrt x}=\\\displaystyle\lim_{x\to 4}\frac{2\sqrt x(x-4)(2+\sqrt x)}{(2-\sqrt x)(2+\sqrt x)}=\displaystyle\lim_{x\to 4}\frac{2\sqrt x(x\!-\!4)(2+\!\sqrt x)}{4+x}\!\\=\!-2\displaystyle\lim_{x\to 4}\left(\sqrt x(2+\!\sqrt x)\right)\!=\!-2(2(2\!+\!2))=-2.8=-16$
2^de manier : (met de regel van de l'Hospital)
$\displaystyle\lim_{x\to 4}\frac{x-4}{\frac{1}{\sqrt x}-\frac{1}{2}}\left(=\frac{0}{0}\right)\buildrel H\over={\displaystyle\lim_{x\to 4}\frac{1-0}{-\frac{1}{x}.\frac{1}{2\sqrt x}-0}=\frac{1}{-\frac{1}{4}.\frac{1}{2.2}}=-16$