UPSC NDA I 2022 Mathematics Solved Paper

Concept:L'Hôpital's rule is used to evaluate limits of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ by differentiating the numerator and denominator separately.Explanation:Let $L = \lim\limits_{x \to 1}\frac{x^3 - 1}{\sqrt x - 1}$.At $x = 1$, the numerator becomes $1^3 - 1 = 0$ and the denominator becomes $\sqrt 1 - 1 = 0$, giving an indeterminate $\frac{0}{0}$ form.Apply L'Hôpital's rule: $L = \lim\limits_{x \to 1} \frac{\frac{d}{dx}(x^3 - 1)}{\frac{d}{dx}(\sqrt x - 1)}$.Differentiate the numerator: $\frac{d}{dx}(x^3 - 1) = 3x^2$.Differentiate the denominator: $\frac{d}{dx}(\sqrt x - 1) = \frac{1}{2\sqrt x}$.Thus $L = \lim\limits_{x \to 1} \frac{3x^2}{\frac{1}{2\sqrt x}} = \lim\limits_{x \to 1} 6x^{5/2}$.Now evaluate the limit: $6 \cdot (1)^{5/2} = 6$.Answer:$6$ (Option C).