Evaluating a 0/0 limit using L'Hopital's Rule

Substituting $x=0$ gives $\frac{0-\sin(0)}{0^3} = \frac{0}{0}$, so the form is indeterminate and L'Hopital's Rule applies. Differentiating numerator and denominator separately gives $\frac{1-\cos(x)}{3x^2}$. Substituting $x=0$ again gives $\frac{1-1}{0} = \frac{0}{0}$, still indeterminate, so we apply L'Hopital's Rule again.

Differentiating again gives $\frac{\sin(x)}{6x}$. Substituting $x=0$ gives $\frac{0}{0}$ once more, so we apply L'Hopital's Rule a third time, differentiating to get $\frac{\cos(x)}{6}$. Now substituting $x=0$ gives $\frac{1}{6}$, which is not indeterminate, so the limit equals $\frac{1}{6}$.