<p>\(\lim_{x \to 1^-} \dfrac{\sqrt{\pi} - \sqrt{2\sin^{-1}x}}{\sqrt{1-x}}\) is equal to __________ (up to four decimal places).</p>
Step-by-Step Solution
Key Concept: As x→1⁻, both numerator and denominator approach 0, creating a 0/0 indeterminate form. Use Taylor expansion of sin⁻¹(x) near x=1 and rationalize the numerator to resolve it.
<p><strong>Step 1:</strong> Verify the indeterminate form. At x=1⁻: numerator = √π - √(π/2) ≠ 0. Check again: sin⁻¹(1) = π/2, so numerator = √π - √(π/2) and denominator = 0. Actually this requires careful analysis.</p><p><strong>Step 2:</strong> Rationalize the numerator by multiplying by conjugate:</p><p>$$\lim_{x \to 1^-} \frac{\sqrt{\pi} - \sqrt{2\sin^{-1}x}}{\sqrt{1-x}} \cdot \frac{\sqrt{\pi} + \sqrt{2\sin^{-1}x}}{\sqrt{\pi} + \sqrt{2\sin^{-1}x}}$$</p><p>$$= \lim_{x \to 1^-} \frac{\pi - 2\sin^{-1}x}{\sqrt{1-x}(\sqrt{\pi} + \sqrt{2\sin^{-1}x})}$$</p><p><strong>Step 3:</strong> Use Taylor expansion near x=1. Let sin⁻¹(x) = π/2 - sin⁻¹(√(1-x)) for x near 1. More directly: as x→1⁻, sin⁻¹(x) = π/2 - √(2(1-x)) + O((1-x)^(3/2))</p><p><strong>Step 4:</strong> Substitute: π - 2sin⁻¹(x) = π - 2[π/2 - √(2(1-x)) + ...] = 2√(2(1-x)) + O((1-x)^(3/2))</p><p>$$\lim_{x \to 1^-} \frac{2\sqrt{2(1-x)}}{\sqrt{1-x}(\sqrt{\pi} + \sqrt{\pi})} = \frac{2\sqrt{2}}{2\sqrt{\pi}} = \sqrt{\frac{2}{\pi}}$$</p><p>∴ Answer: <strong>0.7979</strong></p>
Correct Answer: 0