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Calculus

JEE Main level, single correct, medium

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Question 1 of 4

The value of $\lim_{x \to 0} \frac{e^{2x} - 1}{\sin 3x}$ is equal to:

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Answer: (C) $2/3$

Applying L'Hopital's Rule:

\lim_{x \to 0} \frac{2e^{2x}}{3\cos 3x} = \frac{2(1)}{3(1)} = \frac{2}{3}

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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JEE Main Mathematics practice question on Calculus, with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.

The value of lim⁡x→0e2x−1sin⁡3x\lim_{x \to 0} \frac{e^{2x} - 1}{\sin 3x}limx→0​sin3xe2x−1​ is equal to:

JEE Main question paper PDF

More Mathematics practice

The value of $\lim_{x \to 0} \frac{e^{2x} - 1}{\sin 3x}$ is equal to: