MHTCET 16 May 2024 Shift 1 Paper

Concept:The limit is split into two parts; the second part is evaluated using the logarithmic method and L'Hôpital's rule.Explanation:The given limit is $\lim_{x \to 0^+} \left((\sin x)^{1/x} + \left(\frac{1}{x}\right)^{\sin x}\right)$.First, evaluate $\lim_{x \to 0^+} (\sin x)^{1/x}$.Since $\sin x \to 0$ while $1/x \to \infty$, this is of the form $0^\infty$, so the limit is $0$.Now let $l = \lim_{x \to 0^+} \left(\frac{1}{x}\right)^{\sin x}$.Taking logarithm on both sides,$\log l = \lim_{x \to 0^+} \sin x \log\left(\frac{1}{x}\right) = \lim_{x \to 0^+} (-\sin x \log x)$.Rewrite this as $-\lim_{x \to 0^+} \frac{\log x}{\csc x}$ and apply L'Hôpital's rule.This gives $\log l = \lim_{x \to 0^+} \frac{\sin^2 x}{x \cos x} = \lim_{x \to 0^+} \frac{\tan x}{x} \cdot \sin x = 1 \cdot 0 = 0$.Thus $l = e^0 = 1$.Therefore, the required limit is $0 + 1 = 1$.Answer:Option C. 1