Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 2} \: \frac {x-2} {\frac {1}{x^2} - \frac 14}$ is gelijk aan	A. − 16
B. − 4
C. 2
D. 4
E. 6

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 2} \: \frac {x-2} {\frac {1}{x^2} - \frac 14}$ is equal to	A. − 16
	B. − 4
	C. 2
	D. 4
	E. 6

1^ste manier :
$\begin{aligned} & \displaystyle\lim_{x\to 2}\frac{x-2}{\frac{1}{x^2}-\frac{1}{4}} =\lim_{x\to 2}\frac{x-2}{\frac{4-x^2}{4x^2}} =\lim_{x\to 2}\frac{4x^2\left(x-2\right)}{4-x^2}\\ & =\lim_{x\to 2}\frac{4x^2\left(x-2\right)}{\left(2-x\right)\left(2+x\right)}=\lim_{x\to 2}\frac{-4x^2}{\left(2+x\right)} =\frac{-4.4}{4}=-4 \end{aligned}$
2^de manier : met de regel van de l'Hospital
$\\ \displaystyle\lim_{x\to 2}\frac{x-2}{\frac{1}{x^2}-\frac{1}{4}} =\lim_{x\to 2}\frac{x-2}{x^{-2}-\frac14}\left(=\frac00\right) \overset{H}{=}\lim_{x\to 2}\frac{1-0}{-2x^{-3}-0}=\frac{1}{-2.2^{-3}} =-\frac{2^3}{2}=-4$

^{gricha - v7135 - 29.11.13}