Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 0} \: \frac {\sqrt[3]{x+1}-1} {x}$ is gelijk aan	A. 0
B. 1
C. 2
D. 3
E.

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 0} \: \frac {\sqrt[3]{x+1}-1} {x}$ is equal to	A. 0
B. 1
C. 2
D. 3
E.

1^ste manier : zonder de regel van de l'Hospital
wel rekening houdend met (A − B)(A² + AB + B²) = A³ − B³
$\\\lim_{x\to0}\frac{\sqrt[3]{x+1}-1}{x}=\lim_{x\to0}\frac{\left(\sqrt[3]{x+1}-1\right)\left(\sqrt[3]{\left(x+1\right)^2}+\sqrt[3]{x+1}+1\right)}{x\left(\sqrt[3]{\left(x+1\right)^2}+\sqrt[3]{x+1}+1\right)}\\=\lim_{x\to0}\frac{\left(\sqrt[3]{x+1}\right)^3-1^3}{x\left(\sqrt[3]{\left(x+1\right)^2}+\sqrt[3]{x+1}+1\right)}=\lim_{x\to0}\;{x+1-1}{x\left(\sqrt[3]{\left(x+1\right)^2}+\sqrt[3]{x+1}+1\right)}\\=\lim_{x\to0}\;\frac{1}{\sqrt[3]{\left(x+1\right)^2}+\sqrt[3]{x+1}+1}=\frac{1}{1+1+1}=\frac{1}{3}$
2^de manier :
$\\\lim_{x\to0}\frac{\sqrt[3]{x+1}-1}{x}=\lim_{x\to0}\frac{(x+1)^\frac{1}{3}-1}{x}\left(=\frac{0}{0}\right)\\\buildrel H\over=\lim_{x\to0}\frac{\frac{1}{3}(x+1)^{-\;\frac{2}{3}}}{1}=\frac{1}{3}(0+1)^{-\;\frac{2}{3}}=\frac{1}{3}$