Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

De limiet $\displaystyle \lim_{x \to 0}\; \frac{\sqrt {4+x}-\sqrt {4-x} }{x}$ is gelijk aan	A. 1
B. 2
C. 0
D.
E.

The limit $\displaystyle \lim_{x \to 0}\; \frac{\sqrt {4+x}-\sqrt {4-x} }{x}$ is equal to	A. 1
B. 2
C. 0
D.
E.

1^ste manier : zonder de regel van de l'Hospital
\(\displaystyle\lim_{x\to0}\frac{\sqrt{4+x}-\sqrt{4-x}}{x}=\displaystyle\lim_{x\to0}\frac{\left(\sqrt{4+x}-\sqrt{4-x}\right)\left(\sqrt{4+x}+\sqrt{4-x}\right)}{x\left(\sqrt{4+x}+\sqrt{4-x}\right)}\\=\displaystyle\lim_{x\to0}\frac{\left(4+x\right)-\left(4-x\right)}{x\left(\sqrt{4+x}+\sqrt{4-x}\right)}\;=\displaystyle\lim_{x\to0}\frac{2x}{x\left(\sqrt{4+x}+\sqrt{4-x}\right)}\\=\displaystyle\lim_{x\to0}\frac{2}{\left(\sqrt{4+x}+\sqrt{4-x}\right)}=\frac{2}{\sqrt4+\sqrt4}=\ldots \)
2^de manier : met de regel van de l'Hospital
\(\displaystyle\lim_{x\to0}\frac{\sqrt{4+x}-\sqrt{4-x}}{x}\left(=\frac{0}{0}\right)\buildrel H\over=\lim_{x\to0}\frac{\frac{1}{2\sqrt{4+x}}-\frac{1}{2\sqrt{4-x}}\left(-1\right)}{1}\\=\displaystyle\lim_{x\to0}\left(\frac{1}{2\sqrt{4+x}}+\frac{1}{2\sqrt{4-x}}\right)\;=\frac{1}{2\sqrt4}+\frac{1}{2\sqrt4}=\frac{1}{4}+\frac{1}{4}=\ldots \)