Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to -2}\: \frac{\sqrt{1+\sqrt{x+6}}-\sqrt 3}{x+2}=$	A. 1
B. 3
C. ${\color{DarkBlue} \frac{\sqrt{3}}{24}}$
D.
E.

$\newline {\color{DarkGreen} 1^{ste} manier :}\newline lim_{x \to -2}\frac{\sqrt{1+\sqrt{x+6}}-\sqrt{3}}{x+2}\left (=\frac{0}{0} \right ) \newline = \lim_{x \to -2}\frac{1+\sqrt{x+6}-3}{(x+2)(\sqrt{1+\sqrt{x+6}}+\sqrt{3})} \newline =\lim_{x \to -2}\frac{(\sqrt{x+6}-2)(\sqrt{x+6}+2)}{(x+2)(\sqrt{1+\sqrt{x+6}}+\sqrt{3}) ((\sqrt{x+6} +2) } \newline =\lim_{x \to -2}\frac{x+6-4}{(x+2)(\sqrt{1+\sqrt{x+6}}+\sqrt{3}) (\sqrt{x+6} +2) } \newline =\lim_{x \to -2}\frac{1}{(\sqrt{1+\sqrt{x+6}}+\sqrt{3}) (\sqrt{x+6} +2) } \newline =\frac{1}{(\sqrt{1+2}+\sqrt{3(2+2})}=\frac{1}{2\sqrt{3}.4}=\frac{1}{8\sqrt{3}}=\frac{\sqrt{3}}{24} \newline {\color{DarkGreen} 2^{de} manier :}\newline lim_{x \to -2}\frac{\sqrt{1+\sqrt{x+6}}-\sqrt{3}}{x+2}\left (=\frac{0}{0} \right ) \newline \begin{matrix} H\\ = \end{matrix} \lim_{x \to -2}\frac{1}{2\sqrt{1+\sqrt{x+6}}}\cdot \frac{1}{2\sqrt{x+6}}=\frac{1}{8\sqrt{3}}=\frac{\sqrt{3}}{24}$