$$I=\lim _{x \rightarrow 0} \sin \left(\frac{\mathrm{e}^{x}-x-1-\frac{x^{2}}{2}}{x^{2}}\right)$$
Note that $\sin$ is a continuous function. Thus we have
$$I=\lim _{x \rightarrow 0} \sin \left(\frac{\mathrm{e}^{x}-x-1-\frac{x^{2}}{2}}{x^{2}}\right)=\sin \left(\lim _{x \rightarrow 0} \frac{\mathrm{e}^{x}-x-1-\frac{x^{2}}{2}}{x^{2}}\right)$$
Now we calculate the limit by using L'Hopital Rule
$$\begin{aligned}\lim _{x \rightarrow 0} \frac{\mathrm{e}^{x}-x-1-\frac{x^{2}}{2}}{x^{2}}& = \lim _{x \rightarrow 0} \frac{\mathrm{e}^{x}-1-x}{2x} && \left[ \text{Again $\frac{0}{0}$ form}\right] \\ & = \lim _{x \rightarrow 0} \frac{\mathrm{e}^{x}-1}{2} \\ &=0\end{aligned}$$
Therfore, we get $$I=\sin \left(\lim _{x \rightarrow 0} \frac{\mathrm{e}^{x}-x-1-\frac{x^{2}}{2}}{x^{2}}\right)=\sin\left(0\right)=0.$$
Hence the correct option is (C).