UPSC NDA II 2015 Mathematics Paper

Concept:L'Hospital's Rule is used when a limit gives an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$.For such cases we differentiate the numerator and denominator separately and then take the limit.Explanation:We are given $f(x) = \frac{\sin(e^{x-2} - 1)}{\ln(x-1)}$ and need $\lim_{x \to 2} f(x)$.Direct substitution at $x=2$ gives $\frac{\sin(e^{0} - 1)}{\ln(1)} = \frac{\sin(0)}{0} = \frac{0}{0}$, an indeterminate form.Apply L'Hospital's Rule: differentiate numerator and denominator separately.Numerator derivative: $\frac{d}{dx} \sin(e^{x-2} - 1) = \cos(e^{x-2} - 1) \cdot e^{x-2}$.Denominator derivative: $\frac{d}{dx} \ln(x-1) = \frac{1}{x-1}$.Thus, $\lim_{x \to 2} \frac{\sin(e^{x-2} - 1)}{\ln(x-1)} = \lim_{x \to 2} \frac{\cos(e^{x-2} - 1) \cdot e^{x-2}}{\frac{1}{x-1}}$.Simplify by substituting $x=2$: numerator becomes $\cos(e^{0} - 1) \cdot e^{0} = \cos(0) \cdot 1 = 1$, denominator becomes $\frac{1}{2-1} = 1$.So the limit is $\frac{1}{1} = 1$.Answer:The value of the limit is $1$, which corresponds to option D.