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

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

Substituting $x=0$ into this new expression gives $\frac{1-1}{0} = \frac{0}{0}$, still indeterminate, so L'Hopital's Rule is applied again. The numerator is $\frac{1}{1+x^2} - 1 = \frac{-x^2}{1+x^2}$ and the denominator is $3x^2$, so we need to differentiate both separately. The derivative of the numerator $\frac{-x^2}{1+x^2}$ is $\frac{-2x(1+x^2) - (-x^2)(2x)}{(1+x^2)^2} = \frac{-2x - 2x^3 + 2x^3}{(1+x^2)^2} = \frac{-2x}{(1+x^2)^2}$. The derivative of the denominator $3x^2$ is $6x$. This gives $\lim_{x\to 0} \frac{\frac{-2x}{(1+x^2)^2}}{6x} = \lim_{x\to 0} \frac{-2x}{6x(1+x^2)^2}.$

Since the factor of $x$ appears in both numerator and denominator and $x \to 0$ but $x \neq 0$ during the limit process, we can cancel to obtain $\lim_{x\to 0} \frac{-2}{6(1+x^2)^2}.$ Now direct substitution gives $\frac{-2}{6(1+0)^2} = \frac{-2}{6} = -\frac{1}{3}.$ Therefore, the limit equals $-\frac{1}{3}$.