Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110} \displaystyle \lim_{x\to1}\left ( \frac{x^4+1}{x^2+1} \right )+ \displaystyle \lim_{x\to1}\left ( \frac{x^4-1}{x^2+1} \right )+ \displaystyle \lim_{x\to1}\left ( \frac{x^4-1}{x^2-1} \right )$ is gelijk aan	A. 1
B. 2
C. 3
D. 4
E. 5

$https://latex.codecogs.com/png.image?\dpi{110} \displaystyle \lim_{x\to1}\left ( \frac{x^4+1}{x^2+1} \right )+ \displaystyle \lim_{x\to1}\left ( \frac{x^4-1}{x^2+1} \right )+ \displaystyle \lim_{x\to1}\left ( \frac{x^4-1}{x^2-1} \right )$ is equal to	A. 1
	B. 2
	C. 3
	D. 4
	E. 5

$https://latex.codecogs.com/png.image?\dpi{110} \displaystyle \lim_{x\to1}\left ( \frac{x^4+1}{x^2+1} \right )+ \displaystyle \lim_{x\to1}\left ( \frac{x^4-1}{x^2+1} \right )+ \displaystyle \lim_{x\to1}\left ( \frac{x^4-1}{x^2-1} \right ) = L$
$https://latex.codecogs.com/png.image?\dpi{110}\\{\color{DarkGreen} 1^{ste} \;manier : }\\ L=\frac{1+1}{1+1}+\frac{1-0}{1+1}+\displaystyle \lim_{x\to 1}\frac{(x^2-1)(x^2+1)}{x^2-1} =1+0+(1^2+1)=\cdots \\\\{\color{DarkGreen} 2^{de} \;manier : \text{ (met de regel van de l'Hospital in de tweede limiet)}} \\L= \displaystyle \lim_{x\to 1}\frac{2x^4}{x^2+1}+\displaystyle \lim_{x\to 1}\frac{4x^3}{2x}=\frac 22 + 2 = 3$
${\color{DarkGreen}2^{de} \;manier : \text{ (met de regel van de l'Hospital in de tweede limiet)}}$
$https://latex.codecogs.com/png.image?\dpi{110}\\{\color{DarkGreen} 1^{ste} \;manier : }$