<p>Given <br>\(\lim_{x \to 1^-} \dfrac{\sqrt{\pi} - \sqrt{2\sin^{-1}x}}{\sqrt{1-x}}\)</p><p>Find the value of the limit.</p>
Step-by-Step Solution
Key Concept: Recognize this is an indeterminate form 0/0 that requires L'Hôpital's rule or Taylor expansion. Use the substitution u = sin⁻¹(x) and expand √(π - √(2u)) around u = π/2 as x → 1⁻.
<p><strong>Step 1:</strong> Verify the indeterminate form. As x → 1⁻: sin⁻¹(x) → π/2, so √(2sin⁻¹x) → √π, giving numerator → 0. Also, √(1-x) → 0. This is 0/0 form.</p><p><strong>Step 2:</strong> Let u = sin⁻¹(x), so x = sin(u). As x → 1⁻, u → π/2⁻. Also, 1 - x = 1 - sin(u).</p><p><strong>Step 3:</strong> Rewrite the limit: $$\lim_{u \to \pi/2^-} \frac{\sqrt{\pi} - \sqrt{2u}}{\sqrt{1-\sin(u)}}$$</p><p><strong>Step 4:</strong> Near u = π/2, let h = π/2 - u where h → 0⁺. Then u = π/2 - h and sin(u) = sin(π/2 - h) = cos(h) = 1 - h²/2 + O(h⁴).</p><p><strong>Step 5:</strong> So 1 - sin(u) = h²/2 + O(h⁴), giving √(1-sin(u)) = √(h²/2) = h/√2 + O(h³).</p><p><strong>Step 6:</strong> For the numerator: √(2u) = √(2(π/2 - h)) = √(π - 2h). Using √(π - 2h) = √π·√(1 - 2h/π) = √π(1 - h/π + O(h²)).</p><p><strong>Step 7:</strong> Therefore: √π - √(2u) = √π - √π(1 - h/π + O(h²)) = √π·(h/π + O(h²)) = h/√π + O(h²).</p><p><strong>Step 8:</strong> The limit becomes: $$\lim_{h \to 0^+} \frac{h/\sqrt{\pi}}{h/\sqrt{2}} = \frac{h/\sqrt{\pi}}{h/\sqrt{2}} = \frac{\sqrt{2}}{\sqrt{\pi}} = \sqrt{\frac{2}{\pi}}$$</p><p><strong>Step 9:</strong> Calculate: √(2/π) = √(2/3.14159...) = √(0.63662...) = 0.7977.</p><p><strong>∴ Answer:</strong> 0.7977</p>
Correct Answer: 0.7977