## Calculus (3rd Edition)

$$-6$$
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 $$\lim _{x \rightarrow 0} \frac{6}{-\cos x }=-6.$$