UPSC NDA I 2017 Mathematics Solved Paper

Concept:When a limit yields the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, we can apply L'Hôpital's Rule repeatedly until the limit becomes finite and evaluable.Explanation:We are asked to find $\underset{x \to 0}{\lim}\frac{e^{x} - (1+x)}{x^{2}}$.Direct substitution of $x = 0$ gives $\frac{1 - 1}{0} = \frac{0}{0}$, which is indeterminate.Applying L'Hôpital's Rule: differentiate the numerator and denominator separately.The derivative of the numerator $e^{x} - 1 - x$ is $e^{x} - 1$.The derivative of the denominator $x^{2}$ is $2x$.Thus the limit becomes $\underset{x \to 0}{\lim}\frac{e^{x} - 1}{2x}$, which is still $\frac{0}{0}$.Apply L'Hôpital's Rule again.The derivative of $e^{x} - 1$ is $e^{x}$.The derivative of $2x$ is $2$.Now we have $\underset{x \to 0}{\lim}\frac{e^{x}}{2} = \frac{e^{0}}{2} = \frac{1}{2}$.This value is finite and independent of the original indeterminate form.Answer:$\frac{1}{2}$ (Option B)