UPSC NDA I and II 2020 Math Solved Paper

Concept:Use L’Hôpital’s Rule when a limit gives the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$: differentiate the numerator and denominator separately, then take the limit again.Explanation:First, substitute $x = 1$ into the expression $\frac{x + x^{2} + x^{3} - 3}{x - 1}$.The numerator becomes $1 + 1 + 1 - 3 = 0$, and the denominator becomes $1 - 1 = 0$.So the limit is of the form $\frac{0}{0}$, which allows us to apply L’Hôpital’s Rule.Differentiate the numerator: $\frac{d}{dx}(x + x^2 + x^3 - 3) = 1 + 2x + 3x^2$.Differentiate the denominator: $\frac{d}{dx}(x - 1) = 1$.Now the limit becomes $\underset{x \to 1}{\lim} \frac{1 + 2x + 3x^2}{1}$.Substitute $x = 1$: $1 + 2(1) + 3(1)^2 = 1 + 2 + 3 = 6$.Answer:The limit equals $6$. (Option D)