$\lim _{\mathrm{x} \rightarrow 0} \frac{\mathrm{e}^{2|\sin \mathrm{x}|}-2|\sin \mathrm{x}|-1}{\mathrm{x}^{2}}$

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Limits

lim_{x \to 0} \frac{e ^{2∣ s i n x ∣} - 2∣ s i n x ∣ - 1}{x ^{2}}

Answer: (D)

x \to 0 lim \frac{e ^{2∣ s i n x ∣} - 2∣ sin x ∣ - 1}{x ^{2}}

Let

t = ∣ sin x ∣ \to 0^{+}

x \to 0

, and note

t^{2} / x^{2} \to 1

x \to 0 lim \frac{e ^{2 t} - 2 t - 1}{t ^{2}} \cdot \frac{t ^{2}}{x ^{2}} = t \to 0 lim \frac{e ^{2 t} - 2 t - 1}{t ^{2}} \cdot 1

Using L'Hopital's rule twice on

\frac{e ^{2 t} - 2 t - 1}{t ^{2}}

t \to 0 lim \frac{2 e ^{2 t} - 2}{2 t} = t \to 0 lim \frac{4 e ^{2 t}}{2} = 2

So the limit is equal to

2

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Single correct option

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