JEE Main202131 Aug 2021Morning ShiftMathematicsLimitsActual
lim x → 0 sin 2 π cos 4 x x 4 is equal to :
Options
- A2 π 2
- Bπ 2
- C4 π 2
- D4 π
Correct answer
C. 4 π 2
Step-by-step solution
Given that lim x → 0 sin 2 π 1 - sin 2 x 2 x 4 = lim x → 0 sin 2 π 1 + sin 4 x - 2 sin 2 x x 4 = lim x → 0 sin 2 π - π 2 sin 2 x - sin 4 x x 4 = lim x → 0 sin 2 π 2 sin 2 x - sin 4 x x 4 = lim x → 0 sin π 2 sin 2 x - sin 4 x π 2 sin 2 x - sin 4 x 2 · π 2 2 sin 2 x - sin 4 x 2 x 4 When x   →   0   ⇒ 2 sin 2 x   -   sin 4 x   →   0   Let k   =   2 sin 2 x   -   sin 4