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

Substituting $x=0$ gives $\frac{0 - \arctan(0)}{0^3} = \frac{0}{0}$, confirming the indeterminate form, so L'Hopital's Rule applies. Differentiating numerator and denominator separately: the derivative of $x - \arctan(x)$ is $1 - \frac{1}{1+x^2} = \frac{x^2}{1+x^2}$, and the derivative of $x^3$ is $3x^2$. This gives the new limit $\lim_{x\to 0} \frac{x^2/(1+x^2)}{3x^2}.$

Substituting $x=0$ into this expression gives $\frac{0}{0}$, which is still indeterminate. We apply L'Hopital's Rule a second time. The derivative of $\frac{x^2}{1+x^2}$ is $\frac{2x(1+x^2) - x^2(2x)}{(1+x^2)^2} = \frac{2x}{(1+x^2)^2}$, and the derivative of $3x^2$ is $6x$. This gives $\lim_{x\to 0} \frac{2x/(1+x^2)^2}{6x} = \lim_{x\to 0} \frac{2}{6(1+x^2)^2}.$

This last expression is no longer indeterminate, since substituting $x=0$ gives $\frac{2}{6(1+0)^2} = \frac{2}{6} = \frac{1}{3}$, a well-defined value. Therefore $\lim_{x \to 0} \frac{x - \arctan(x)}{x^3} = \frac{1}{3}.$

The rule was applied exactly twice. The first application transformed the original zero over zero form into a new fraction. Substituting x=0 into this new fraction still yielded 0/0, which is indeterminate. Therefore, a second application of L'Hopital's Rule was necessary, differentiating the numerator and denominator again to reach the final non-indeterminate expression 2/[6(1+x²)²], which evaluates to 1/3.