Repeated L'Hopital's Rule on a 0/0 limit

Substituting $x=0$ gives $(\tan(0)-0)/0^3 = 0/0$, so L'Hopital's Rule applies. Differentiating numerator and denominator gives $\frac{\sec^2(x)-1}{3x^2}$. Substituting $x=0$ again gives $(\sec^2(0)-1)/0 = 0/0$, so the rule applies again.

Differentiating again gives $\frac{2\sec^2(x)\tan(x)}{6x}$. Substituting $x=0$ still gives $0/0$, so apply L'Hopital once more, using the product rule on the numerator: the derivative of $2\sec^2(x)\tan(x)$ is $4\sec^2(x)\tan^2(x) + 2\sec^4(x)$, and the derivative of $6x$ is $6$.

This gives $\lim_{x\to 0}\frac{4\sec^2(x)\tan^2(x)+2\sec^4(x)}{6}$. Substituting $x=0$ (where $\sec(0)=1,\ \tan(0)=0$) gives $\frac{0+2}{6} = \frac{1}{3}$.