L'Hôpital's rule, twice

First application. The numerator and denominator vanish at $0$ and are differentiable:
$$\frac{(x-x\cos x)'}{(x-\sin x)'}=\frac{1-\cos x+x\sin x}{1-\cos x}.$$
This is again a $\frac00$ form.

Second application.
$$\begin{aligned} \frac{(1-\cos x+x\sin x)'}{(1-\cos x)'}&=\frac{\sin x+\sin x+x\cos x}{\sin x}\\&=2+\frac x{\sin x}\cos x\longrightarrow2+1\cdot1\\&=3. \end{aligned}$$

Conclusion. The limit is $3$, since each application preserves the limit once the new limit exists.

Check by Taylor. $x-x\cos x=\frac{x^3}2+o(x^3)$ and $x-\sin x=\frac{x^3}6+o(x^3)$, so the quotient tends to $\frac{1/2}{1/6}=3$.