+1 vote
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If $I=\lim _{x \rightarrow 0} \sin \left(\frac{\mathrm{e}^{x}-x-1-\frac{x^{2}}{2}}{x^{2}}\right)$, then limit

(A) does not exist

(B) exists and equals 1

(C) exists and equals 0

(D) exists and equals $\frac{1}{2}$
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$$I=\lim _{x \rightarrow 0} \sin \left(\frac{\mathrm{e}^{x}-x-1-\frac{x^{2}}{2}}{x^{2}}\right)$$

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).

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