Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

De limiet $https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 1} \: \frac{ \sqrt[3]x - 1}{\sqrt x - 1}$ is gelijk aan	A.
B.
C.
D.
E.

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 1} \: \frac{ \sqrt[3]x - 1}{\sqrt x - 1}=$	A.
B.
C.
D.
E.

1^ste manier : zonder de regel van de l'Hospital
$\\\displaystyle\lim_{x\to 1}\:\frac{\sqrt[3]x-1}{\sqrt x-1}=\lim_{x\to 1}\:\frac{(\sqrt[3]x-1)(\sqrt x+1)(\sqrt[3]{x^2}+\sqrt[3]x+1)}{(\sqrt x-1)(\sqrt x+1)(\sqrt[3]{x^2}+\sqrt[3]x+1)}\\=\lim_{x\to 1}\:\frac{(x-1)(\sqrt x+1)}{(x-1)(\sqrt[3]{x^2}+\sqrt[3]x+1)}=\lim_{x\to 1}\:\frac{\sqrt x+1}{\sqrt[3]{x^2}+\sqrt[3]x+1}=\frac{1+1}{1+1+1}=\frac23$
2^de manier : met de regel van de l'Hospital
$\\\displaystyle\lim_{x\to 1}\:\frac{\sqrt[3]x-1}{\sqrt x-1}\left(=\frac00\right)\buildrel H\over=\displaystyle\lim_{x\to 1}\:\frac{x^{\frac13}-1}{x^{\frac12}-1}=\displaystyle\lim_{x\to 1}\:\frac{\frac13 x^{-\frac23}}{\frac12 x^{-\frac12}}=\frac23\displaystyle\lim_{x\to 1}\:\frac{x^{-\frac23}}{x^{-\frac12}}=\frac23.\frac11==\frac23$