<p>Find the value of \( n \) if <br/> \[ \lim_{x \to 0} \frac{\ln(1 + \sin^3 x \cos^2 x)\cot(\ln^3(1+x))\tan^4 x}{\sin(\sqrt{x^2+2} - \sqrt{2}) \cdot \ln(1+x^2)} = \sqrt{n} \]</p>
Step-by-Step Solution
Key Concept: Use Taylor expansions for all transcendental functions around x=0: ln(1+u)≈u, sin(u)≈u, tan(u)≈u, cot(u)≈1/u, and √(2+x²)≈√2 + x²/(2√2). The limit reduces to comparing leading order terms in numerator and denominator.
<p><strong>Step 1: Expand numerator components</strong></p><p>• ln(1 + sin³x cos²x) ≈ sin³x cos²x ≈ x³ (since sin x ≈ x, cos x ≈ 1)</p><p>• ln(1+x) ≈ x, so ln³(1+x) ≈ x³</p><p>• cot(ln³(1+x)) = cot(x³) ≈ 1/x³</p><p>• tan⁴x ≈ x⁴</p><p><strong>Numerator:</strong> x³ · (1/x³) · x⁴ = x⁴</p><p><strong>Step 2: Expand denominator components</strong></p><p>• √(x²+2) - √2 = √2(√(1+x²/2) - 1) ≈ √2 · (x²/4) = x²/(2√2)</p><p>• sin(x²/(2√2)) ≈ x²/(2√2)</p><p>• ln(1+x²) ≈ x²</p><p><strong>Denominator:</strong> [x²/(2√2)] · x² = x⁴/(2√2)</p><p><strong>Step 3: Compute the limit</strong></p><p>∴ lim = x⁴ ÷ [x⁴/(2√2)] = 2√2 = √8</p><p><strong>Therefore n = 8</strong></p>
Correct Answer: 8