Verified Solution • 2025–26 Session moderate QJEEM24Ma9f50: JEE Main 2024 Mathematics - Class 12 JEEM24Ma9f50 Solution: lim x → 0 e 2 limx→0e2|sinx|−2|sinx|−1x2 Multiple Choice Options A is equal to 1 B does not exist C is equal to −1-1−1 D is equal to 2 Step-by-Step Verified Solution amp;limx→0e2∣sinx∣−2∣sinx∣−1x2amp;limx→0e2∣sinx∣−2∣sinx∣−1∣sinx∣2×sin2xx2\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}amp;x→0limx2e2∣sinx∣−2∣sinx∣−1amp;x→0lim∣sinx∣2e2∣sinx∣−2∣sinx∣−1×x2sin2x Let ∣sinx∣=t|\sin \mathrm{x}|=\mathrm{t}∣sinx∣=t amp;limt→0e2t−2t−1t2×limx→0sin2xx2amp;=limt→02e2t−22t×1=2×1=2\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}amp;t→0limt2e2t−2t−1×x→0limx2sin2xamp;=t→0lim2t2e2t−2×1=2×1=2 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring