Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle\lim_{x \to 2} \frac {2x^3-7x^2+4x+4} {x^4-8x^2+16}$ is gelijk aan	A. $\frac{t}{11}$
B. $\frac{5}{16}$
C. $\frac{5}{8}$
D. $\frac{5}{4}$
E. 0

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle\lim_{x \to 2} \frac {2x^3-7x^2+4x+4} {x^4-8x^2+16}$ is equal to	A. $\frac{t}{11}$
	B. $\frac{5}{16}$
	C. $\frac{5}{8}$
	D. $\frac{5}{4}$
	E. 0

1^ste manier : zonder de regel van de l'Hospital
$\\\displaystyle\lim_{x\to 2}\frac{2x^3-7x^2+4x+4}{x^4-8x^2+16}=\displaystyle\lim_{x\to 2}\frac{\left(x-2\right)\left(2x^2-3x-2\right)}{(x^2-4)^2}=\displaystyle\lim_{x\to 2}\frac{2x^2-3x-2}{(x+2)^2(x-2)}\\=\displaystyle\lim_{x\to 2}\frac{\left(x-2\right)\left(2x+1\right)}{(x+2)^2(x-2)}=\displaystyle\lim_{x\to 2}\frac{2x+1}{(x+2)^2}=\frac{5}{16}$
2^de manier : met de regel van de l'Hospital
$\\\displaystyle\lim_{x\to 2}\frac{2x^3-7x^2+4x+4}{x^4-8x^2+16}\left(=\frac{0}{0}\right)\overset{H}{=}\displaystyle\lim_{x\to 2}\frac{6x^2-14x+4}{4x^3-16x}\left(=\frac{0}{0}\right)\\\overset{H}{=}\displaystyle\lim_{x\to 2}\frac{12x-14}{12x^2-16}=\frac{24-14}{48-16}=\frac{10}{32}=\frac{5}{16}$