Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{{x \to 0}} \:\frac {1-\sqrt{1\,-\,\sin x}} {x}$ is gelijk aan	A. 1
B. − 2
C. − 1
D.
E.

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{{x \to 0}} \:\frac {1-\sqrt{1\,-\,\sin x}} {x}$ is equal to	A. 1
	B. − 2
	C. − 1
	D.
	E.

1^ste manier : zonder de regel van de l'Hospital
$\\\displaystyle\lim_{{x\to 0}}\:\frac{1-\sqrt{1\,-\,\sin x}}{x}=\lim_{{x\to 0}}\:\frac{(1-\sqrt{1\,-\,\sin x}).(1+\sqrt{1\,-\,\sin x})}{x.(1+\sqrt{1\,-\,\sin x)}}=\lim_{{x\to 0}}\:\frac{1-1+\sin x}{x.(1+\sqrt{1\,-\,\sin x)}}\\=\lim_{{x\to 0}}\:\frac{\sin x}{x}.\lim_{{x\to 0}}\:\frac{1}{1+\sqrt{1\,+\,\sin x}}=1.\frac12=\frac12$
2^de manier : met de regel van de l' Hospital
$\\\displaystyle\lim_{{x\to 0}}\:\frac{1-\sqrt{1\,-\,\sin x}}{x}\left(=\frac00\right)\buildrel H\over=\lim_{{x\to 0}}\:\frac{1.(-\cos x)}{2\sqrt{1-\sin x}}=\frac12$