[tex]\displaystyle\lim_{x\to 0}\sqrt[3]{\frac{x-\sin x}{x^3}}=\displaystyle\sqrt[3]{\lim_{x\to 0}\frac{x-\sin x}{x^3}}[/tex]
Pentru limita de sub radical se aplică l'Hospital de două ori:
[tex]\displaystyle\lim_{x\to 0}\frac{x-\sin x}{x^3}=\lim_{x\to 0}\frac{1-\cos x}{3x^2}=\\=\displaystyle\lim_{x\to 0}\frac{\sin x}{6x}=\displaystyle\frac{1}{6}\lim_{x\to 0}\frac{\sin x}{x}=\frac{1}{6}[/tex]
Deci limita este [tex]\displaystyle\frac{1}{\sqrt[3]{6}}[/tex]
