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

Substituting $x=0$ gives $\frac{\tan(0)-0}{0^3} = \frac{0}{0}$, an indeterminate form, so L'Hopital's Rule applies. Differentiating numerator and denominator separately: $\frac{d}{dx}[\tan(x)-x] = \sec^2(x)-1$ and $\frac{d}{dx}[x^3] = 3x^2$. This gives the new limit $\lim_{x\to 0} \frac{\sec^2(x)-1}{3x^2}$, which again equals $\frac{0}{0}$ since $\sec^2(0)-1=0$.

Applying L'Hopital's Rule again: $\frac{d}{dx}[\sec^2(x)-1] = 2\sec^2(x)\tan(x)$ and $\frac{d}{dx}[3x^2] = 6x$. The limit becomes $\lim_{x\to 0} \frac{2\sec^2(x)\tan(x)}{6x}$, still $\frac{0}{0}$ since $\tan(0)=0$. Applying L'Hopital's Rule a third time, differentiate numerator using the product rule: $\frac{d}{dx}[2\sec^2(x)\tan(x)] = 4\sec^2(x)\tan^2(x) + 2\sec^4(x)$, and the denominator gives $6$. Substituting $x=0$: $\frac{4(1)(0)+2(1)}{6} = \frac{2}{6} = \frac{1}{3}$.

The rule was applied three times total. The first application turned the original zero over zero form into a new zero over zero form involving secant squared minus one over three x squared. The second application produced another zero over zero form involving secant squared times tangent over six x. The third application finally removed the indeterminate form, giving the final value one third.