sin x − x cos x over x cubed

Idea. Taylor expansions replace functions by polynomials plus a negligible remainder. The question is how far to expand: the denominator is $x^3$, so we need the numerator up to $x^3$.

Expansions. As $x\to0$,
$$\sin x=x-\frac{x^3}6+o(x^3),\qquad \cos x=1-\frac{x^2}2+o(x^2).$$
Multiplying the second one by $x$,
$$x\cos x=x-\frac{x^3}2+o(x^3).$$

Computation.
$$\sin x-x\cos x=\left(-\frac16+\frac12\right)x^3+o(x^3)=\frac{x^3}3+o(x^3).$$
Dividing by $x^3$ gives $\frac13+o(1)$, so the limit is $\dfrac13$.

Rule of thumb. Expand numerator and denominator until the first term that does not cancel appears. If you stop too early (at order $1$ here), you get $0-0$ and learn nothing.