Limits & Continuity: lim_xto0frace^2|sin x|-2|sin x|-1x^2 | JEE Main 2024 | Yukti Vault

Yukti Vault

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

is equal to $2$✓ Correct answeris equal to $1$is equal to $-1$does not exist

Solution

Let $s=|\sin x|$. $e^{2s}-2s-1=\frac{(2s)^2}{2}+\dots=2s^2+O(s^3)$. Since $s^2/x^2=\sin^2x/x^2\to1$, the limit is $2$ from both sides. Option (A).

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