WBJEE 2021 - Question 1 - Find the limit of I.

Question

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 $\frac{1}{2}$

Answer 1

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

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).