28.60. Calculate the limit $\lim _{x \rightarrow 0} \frac{x-\sin x}{x^{3}}$ using L'Hôpital's rule.
28.60. Applying L'Hôpital's rule twice, we get

$$
\lim _{x \rightarrow 0} \frac{x-\sin x}{x^{3}}=\lim _{x \rightarrow 0} \frac{1-\cos x}{3 x^{2}}=\lim _{x \rightarrow 0} \frac{\sin x}{6 x}=\frac{1}{6}
$$