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

Substituting $x=0$ into $\frac{\arcsin(x)-x}{x^3}$ gives $\frac{0-0}{0} = \frac{0}{0}$, so L'Hopital's Rule applies. Differentiating numerator and denominator separately: the derivative of $\arcsin(x)-x$ is $\frac{1}{\sqrt{1-x^2}} - 1$, and the derivative of $x^3$ is $3x^2$. This gives the new limit $\lim_{x\to 0} \frac{\frac{1}{\sqrt{1-x^2}} - 1}{3x^2}.$ Substituting $x=0$ again gives $\frac{1-1}{0} = \frac{0}{0}$, so the form is still indeterminate and another application of L'Hopital's Rule is justified.

Differentiate again: the derivative of $\frac{1}{\sqrt{1-x^2}} - 1 = (1-x^2)^{-1/2} - 1$ is $x(1-x^2)^{-3/2}$, and the derivative of $3x^2$ is $6x$. This gives $\lim_{x\to 0} \frac{x(1-x^2)^{-3/2}}{6x} = \lim_{x\to 0} \frac{(1-x^2)^{-3/2}}{6}.$ Now substituting $x=0$ directly (no longer indeterminate) gives $\frac{1}{6}$, so the original limit equals $\frac{1}{6}$.