Answer
$$-6$$
Work Step by Step
Since we have
$$
\lim _{x \rightarrow 0} \frac{x^3}{\sin x -x}=\frac{0}{0}
$$
then we can apply L’Hôpital’s Rule as follows
$$
\lim _{x \rightarrow 0} \frac{3x^2}{\cos x -1}=\frac{0}{0}.
$$
We have to apply L’Hôpital’s Rule again
$$
\lim _{x \rightarrow 0} \frac{6x}{-\sin x }=\frac{0}{0}.
$$
Since the form is intermediate, we have to apply L’Hôpital’s Rule as follows
$$
\lim _{x \rightarrow 0} \frac{6}{-\cos x }=-6.
$$
