Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 0} \: \frac{\sqrt{4+x}+\sqrt{4-x}-4}{x}$ is gelijk aan	A. 0
B. 1
C. 2
D. 4
E.

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 0} \: \frac{\sqrt{4+x}+\sqrt{4-x}-4}{x}$ is equal to	A. 0
B. 1
C. 2
D. 4
E. ∞

1^ste manier : met de regel van de l'Hospital
$\\\displaystyle\lim_{x\to 0}\frac{\sqrt{4+x}+\sqrt{4-x}-4}{x}\left(=\frac00\right)\overset{H}{=}\displaystyle\lim_{x\to 0}\:\left(\frac{1}{2\sqrt{4+x}}-\frac{1}{2\sqrt{4\!-\!x}}\right)\!=\!\frac{1}{2\sqrt4}-\frac{1}{2\sqrt4}=0$
2^de manier : zonder de regel van de l'Hospital
$\\\displaystyle\lim_{x\to 0}\frac{\sqrt{4+x}+\sqrt{4-x}-4}{x}=\displaystyle\lim_{x\to 0}\frac{(\sqrt{4+x}+\sqrt{4-x}-4)(\sqrt{4+x}+\sqrt{4-x}+4)}{x.(\sqrt{4+x}+\sqrt{4-x}+4)}=\\\displaystyle\lim_{x\to 0}\frac{(\sqrt{4+x}+\sqrt{4-x})^2-16}{x.(\cdots)}=\displaystyle\lim_{x\to 0}\frac{4+x+2\sqrt{16-x^2}+4-x-16}{x.(\cdots)}=$
$\\=\displaystyle\lim_{x\to 0}\frac{(2\sqrt{16-x^2}-8)(2\sqrt{16-x^2}-8)}{x.(\cdots)(2\sqrt{16-x^2}-8))}=\displaystyle\lim_{x\to 0}\frac{4(16-x^2)-64}{x.(\cdots)(\cdots)}=\displaystyle\lim_{x\to 0}\frac{-4x^2}{x.(\cdots)(\cdots)}\\=\displaystyle\lim_{x\to 0}\frac{-4x}{(\cdots)(\cdots)}=\frac{0}{(\sqrt4+\sqrt4+4)(2.4+8)}=0$