How can I evaluate this limit without using L'hospital, derivatives or series expansion? I have troubles solving this particular limit without using series expansion, L'Hospital rule or derivatives.
$$\lim_{x \to 0}{\frac{x\sin{x}+2(\cos{x}-1)}{x^3}}$$
I have tried multiplying by $(\cos{x}+1)$ and got rid of one $x$ in the denominator, but got arguably something even messier. I got this:
$$\lim_{x \to 0}{\frac{x(\cos{x}+1)-2\sin{x}}{x^2(\cos{x}+1)}}$$
Any help would be appreciated.

As J.G. already shown it holds $$\begin{align}\frac{x\sin x+2(\cos x-1)}{x^3}&=\frac{2x\sin\frac{x}{2}\cos\frac{x}{2}-4\sin^2\frac{x}{2}}{x^3}\\&=\frac{4\sin^2\frac{x}{2}}{x^2}\frac{\frac{x}{2}\cot\frac{x}{2}-1}{x} \\ &= \left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2 \frac{f(x) - f(0)}{x - 0} \end{align}$$
for $f(x) = x\cot(x)$
So we get by the mean value theorem: $$\lim_{x\to 0} \frac{x\sin x+2(\cos x-1)}{x^3} = 0\cdot\lim_{x\to 0} f'(x) = 0\cdot1 = 0$$
