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JEE Main 2019 · 12 Jan · Shift II

JEE Main 2019 Mathematics question on Limits Continuity and Differentiability

lim⁡x→1−π−2sin⁡−1x1−x\lim_{x \to 1^-} \dfrac{\sqrt{\pi} - \sqrt{2\sin^{-1}x}}{\sqrt{1-x}} is equal to
  1. A.
    π2\sqrt{\dfrac{\pi}{2}}
  2. B.
    2π\sqrt{\dfrac{2}{\pi}}
  3. C.
    π\sqrt{\pi}
  4. D.
    12π\dfrac{1}{\sqrt{2\pi}}
Show answer and solution

Correct option: B

Solution

Let θ = sin⁻¹x, so as x → 1⁻, θ → π/2⁻. Write sin⁻¹x = π/2 - ε where ε → 0⁺, and x = sin(π/2 - ε) = cosε ≈ 1 - ε²/2, so 1 - x ≈ ε²/2.

The numerator: √π - √(2sin⁻¹x) = √π - √(π - 2ε) = √π(1 - √(1 - 2ε/π)) ≈ √π · (ε/π) = ε/√π.

The denominator: √(1-x) ≈ √(ε²/2) = ε/√2.

Limit = (ε/√π)/(ε/√2) = √2/√π = √(2/π).

[JEE Main 2019 · 12 Jan · Shift II]

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