Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

De som van de limieten $https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 1}\frac{\sqrt x - 1}{x - 1}\;\;en\;\;\displaystyle \lim_{x \to 1}\frac{x^2 - 1}{x - 1}$ is gelijk aan	A. 1
B. 2,5
C. 0
D. 4
E. geen van de vorige

The sum of the limits $https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 1}\frac{\sqrt x - 1}{x - 1}\;\;en\;\;\displaystyle \lim_{x \to 1}\frac{x^2 - 1}{x - 1}$ is equal to	A. 1
B. 2.5
C. 0
D. 4
E. none of the above

1^ste manier :
$https://latex.codecogs.com/png.image?\dpi{110}\\\displaystyle \lim_{x \to 1}\frac{\sqrt x - 1}{x - 1}\;+\;\displaystyle \lim_{x \to 1}\frac{x^2 - 1}{x - 1}=\displaystyle \lim_{x \to 1}\frac{(\sqrt x-1)(\sqrt x+1)}{(x-1)(\sqrt x+1)}+\displaystyle \lim_{x \to 1}\frac{(x^2-1)(x+1)}{(x-1)(x+1)}\\=\displaystyle \lim_{x \to 1}\frac{1}{\sqrt x + 1}+\displaystyle \lim_{x \to 1}\frac{x+1}{1}=\frac12+2=\frac52$
2^de manier :
$https://latex.codecogs.com/png.image?\dpi{110}\\\displaystyle \lim_{x \to 1}\frac{\sqrt x - 1}{x - 1}\;+\;\displaystyle \lim_{x \to 1}\frac{x^2 - 1}{x - 1}\overset{H}{=}\displaystyle \lim_{x \to 1}\frac{\frac1{2\sqrt x}}{1}+\displaystyle \lim_{x \to 1}\frac{2x}{1}=\frac12+2=\frac52$
3^de manier :
$https://latex.codecogs.com/png.image?\dpi{110}\\\displaystyle \lim_{x \to 1}\frac{\sqrt x - 1}{x - 1}\;+\;\displaystyle \lim_{x \to 1}\frac{x^2 - 1}{x - 1}=\displaystyle \lim_{x \to 1}\frac{\sqrt x+x^2-2}{x-1}=\displaystyle \lim_{x \to 1}\frac{x^2-1+\sqrt x-1}{x-1}=\displaystyle \lim_{x \to 1}\frac{(x-1)(x+1)+(\sqrt x-1)}{x-1}=\displaystyle \lim_{x \to 1}\frac{(\sqrt x-1)(\sqrt x+1)(x+1)+(\sqrt x-1)}{(\sqrt x-1)(\sqrt x+1)}=\displaystyle \lim_{x \to 1}\frac{(\sqrt x+1)(x+1)+1}{\sqrt x+1}=\frac{2.2+1}{2}=\frac52$