JEE Main2014MathematicsLimitsActual
lim x → 0 sin πcos 2 x x 2 is equal to
Options
- A- π
- Bπ
- Cπ 2
- D1
Correct answer
B. π
Step-by-step solution
We have lim x → 0 sin πcos 2 x x 2 = lim x → 0 sin π - πcos 2 x x 2 ∵   sin π - θ = sin  θ = lim x → 0 sin π   sin 2 x π   sin 2 x × π   sin 2 x x 2 = lim x → 0 π sin π   sin 2 x π   sin 2 x sin  x x 2 = π 1 1 As ,   lim x → 0 sin   t t = 1 = π .