Let $ x\in\mathbb{R}^{*} $, observe that : $$ \fbox{$\begin{array}{rcl}\displaystyle\frac{x-\sin{x}}{x^{2}}=\frac{x}{2}\int_{0}^{1}{\left(1-t\right)^{2}\cos{\left(tx\right)}\,\mathrm{d}t}\end{array}$} $$
Using the fact that $ \left(\forall t\in\left[0,1\right]\right),\ \left|\cos{\left(tx\right)}\right|\leq 1 $, we have : \begin{aligned} \left|\frac{x-\sin{x}}{x^{2}}\right|=\frac{\left|x\right|}{2}\left|\int_{0}^{1}{\left(1-t\right)^{2}\cos{\left(tx\right)}\,\mathrm{d}t}\right|&\leq\frac{\left|x\right|}{2}\int_{0}^{1}{\left(1-t\right)^{2}\left|\cos{\left(tx\right)}\right|\mathrm{d}t}\\&\leq\frac{\left|x\right|}{2}\int_{0}^{1}{\left(1-t\right)^{2}\,\mathrm{d}t} \end{aligned}
Which means $ \left(\forall x\in\mathbb{R}^{*}\right),\ \left|\frac{x-\sin{x}}{x^{3}}\right|\leq\frac{\left|x\right|}{6} $, and thus $ \lim\limits_{x\to 0}{\frac{x-\sin{x}}{x^{2}}}=0 \cdot $
Let $ x\in\mathbb{R}^{*} $, observe that : $$ \fbox{$\begin{array}{rcl}\displaystyle\frac{x^{2}+2\left(\cos{x}-1\right)}{x^{3}}=\int_{0}^{1}{\left(1-t\right)^{2}\sin{\left(tx\right)}\,\mathrm{d}t}\end{array}$} $$
Using the fact that $ \left(\forall t\in\left[0,1\right]\right),\ \left|\sin{\left(tx\right)}\right|\leq t\left|x\right| $, we have : \begin{aligned} \left|\frac{x^{2}+2\left(\cos{x}-1\right)}{x^{3}}\right|=\left|\int_{0}^{1}{\left(1-t\right)^{2}\sin{\left(tx\right)}\,\mathrm{d}t}\right|&\leq\int_{0}^{1}{\left(1-t\right)^{2}\left|\sin{\left(tx\right)}\right|\mathrm{d}t}\\&\leq\left|x\right|\int_{0}^{1}{t\left(1-t\right)^{2}\,\mathrm{d}t} \end{aligned}
Which means $ \left(\forall x\in\mathbb{R}^{*}\right),\ \left|\frac{x^{2}+2\left(\cos{x}-1\right)}{x^{3}}\right|\leq\frac{\left|x\right|}{12} $, and thus $ \lim\limits_{x\to 0}{\frac{x^{2}+2\left(\cos{x}-1\right)}{x^{3}}}=0 \cdot $
Hence, $$ \lim_{x\to 0}{\frac{x\sin{x}+2\left(\cos{x}-1\right)}{x^{3}}}=\lim_{x\to 0}{\left(\frac{x^{2}+2\left(\cos{x}-1\right)}{x^{3}}-\frac{x-\sin{x}}{x^{2}}\right)}=0 $$