Răspunsuri

2014-08-03T09:30:55+03:00
\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}}
Pentru limita de sub radical se aplică l'Hospital de două ori:
\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}
Deci limita este \displaystyle\frac{1}{\sqrt[3]{6}}