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