JEE Main201912 Jan 2019Evening ShiftMathematicsLimitsActual
lim x → 1 - π - 2 sin - 1 x 1 - x is equal to
Options
- Aπ
- B2 π
- C1 2 π
- Dπ 2
Correct answer
B. 2 π
Step-by-step solution
Given limit can be written as L = lim x → 1 - π - 2 sin - 1 x 1 - x × π + 2 sin - 1 x π + 2 sin - 1 x ⇒ L = lim x → 1 - π - 2 sin - 1 x 1 - x   π + 2 sin - 1 x = lim x → 1 - 2 cos - 1 x 1 - x   π + 2 sin - 1 x Let K = lim x → 1 - cos - 1 x 1 - x and put x = cos θ , we get K = lim θ → 0 θ 2 . 2 2 . sin θ 2 = lim θ → 0 2 sin θ 2 θ 2 = 2   ∵ lim x → 0 sin x x = 1 ∴   L =