Find $\lim_{x\to 0}\frac{\sin x-x\cos x}{x^3}$ $$\lim_{x\to 0}\frac{\sin x-x\cos x}{x^3}$$ How do I go about doing this? I can see no simple way of doing this. Application of l'Hopital's rule would be very laborious. A Taylor expansion seems feasible but is that the best way? It seems like it may be also very laborious.

Taylor expanding we find that: \begin{equation} \begin{aligned} \lim_{x\to 0}\frac{\sin x-x\cos x}{x^3}&=\lim_{x\to 0}\frac{x-\frac{x^3}{6}+\mathcal{O}\left(x^4\right)-x\left[1-\frac{x^2}{2}+\mathcal{O}\left(x^4\right)\right]}{x^3}\\\ &=\lim_{x\to 0}\frac{x^3\left(\frac{1}{2}-\frac{1}{6}\right)+\mathcal{O}\left(x^4\right)}{x^3}\\\ &=\frac{1}{3} \end{aligned} \end{equation}