Home / JEE Main / Questions / Limits / Question JEE Main 2017 · Mathematics · Limits · easy
Lim x→0(e^3x-1)/(sin 2x) equals Single correct option
lim x → 0 e 3 x − 1 sin 2 x \displaystyle\lim_{x\to0}\dfrac{e^{3x}-1}{\sin 2x} x → 0 lim sin 2 x e 3 x − 1 equals
A. 2 3 \frac{2}{3} 3 2 B. 1 1 1 C. 3 2 \frac{3}{2} 2 3 D. 6 6 6 Show answer and solution Answer: (C) 3 2 \frac{3}{2} 2 3
Write the limit as
e 3 x − 1 3 x ⋅ 2 x sin 2 x ⋅ 3 x 2 x \dfrac{e^{3x}-1}{3x}\cdot\dfrac{2x}{\sin2x}\cdot\dfrac{3x}{2x} 3 x e 3 x − 1 ⋅ sin 2 x 2 x ⋅ 2 x 3 x . The first two factors tend to
1 1 1 by the standard limits
e u − 1 u → 1 \dfrac{e^u-1}{u}\to1 u e u − 1 → 1 and
sin u u → 1 \dfrac{\sin u}{u}\to1 u sin u → 1 , leaving
3 2 \dfrac32 2 3 . L'Hospital's rule gives the same:
3 e 3 x 2 cos 2 x → 3 2 \dfrac{3e^{3x}}{2\cos2x}\to\dfrac32 2 cos 2 x 3 e 3 x → 2 3 .
Original question written in the JEE Main 2017 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake