JEE Main 2024 (Online) 31st January Morning ShiftQuestion 1
$\lim_{x\to0}\frac{e^{2|\sin x|}-2|\sin x|-1}{x^{2}}$
Limits & Continuity
Solution
Concept. Put $u=2|\sin x|\to0$; the modulus disappears after squaring. Formulas used. (1) $e^u-u-1=\frac{u^2}2+O(u^3)$(2) $\lim_{x\to0}\frac{\sin^2x}{x^2}=1$Step 1 - Expand. With $u=2|\sin x|$, by (1): $$e^{u}-u-1=\frac{4\sin^2x}{2}+O(|\sin x|^3)=2\sin^2x+\dots$$ Step 2 - Divide. By (2), $\frac{2\sin^2x}{x^2}\to2$, from both sides (no sign issue since $\sin^2x=|\sin x|^2$). Step 3. The limit exists and equals $2$. Answer: A