Question

($\lim_{x \to \ 0} \sqrt[3]{ \frac{x-sinx}{ x^{3} } }$

Answer 1

$\displaystyle\lim_{x\to 0}\sqrt[3]{\frac{x-\sin x}{x^3}}=\displaystyle\sqrt[3]{\lim_{x\to 0}\frac{x-\sin x}{x^3}}$
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 $\displaystyle\frac{1}{\sqrt[3]{6}}$