JEE MAIN - Mathematics (2024 - 31st January Morning Shift - No. 1)

$$\lim _\limits{x \rightarrow 0} \frac{e^{2|\sin x|}-2|\sin x|-1}{x^2}$$
is equal to 1
does not exist
is equal to $$-1$$
is equal to 2

Explanation

$$\begin{aligned} & \lim _{x \rightarrow 0} \frac{e^{2|\sin x|}-2|\sin x|-1}{x^2} \\ & \lim _{x \rightarrow 0} \frac{e^{2|\sin x|}-2|\sin x|-1}{|\sin x|^2} \times \frac{\sin ^2 x}{x^2} \end{aligned}$$

Let $$|\sin \mathrm{x}|=\mathrm{t}$$

$$\begin{aligned} & \lim _{t \rightarrow 0} \frac{e^{2 t}-2 t-1}{t^2} \times \lim _{x \rightarrow 0} \frac{\sin ^2 x}{x^2} \\ & =\lim _{t \rightarrow 0} \frac{2 e^{2 t}-2}{2 t} \times 1=2 \times 1=2 \end{aligned} $$

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