<p>Evaluate \(\lim_{x \to 0} \dfrac{64^x - 32^x - 16^x + 4^x + 2^x - 1}{\left[\sqrt{(15 + \cos x)} - 4\right]\sin x}\)</p>
Step-by-Step Solution
Key Concept: Express all terms as powers of 2 (64=2^6, 32=2^5, 16=2^4, 4=2^2), then use Taylor expansion: 2^(ax) = 1 + ax·ln2 + (ax·ln2)²/2 + ... to find the numerator's leading behavior. For the denominator, rationalize √(15+cosx) - 4 and use cosx ≈ 1 - x²/2.
<p><strong>Step 1:</strong> Write all bases as powers of 2: 64^x = (2^6)^x = 2^(6x), 32^x = 2^(5x), 16^x = 2^(4x), 4^x = 2^(2x)</p><p><strong>Step 2:</strong> Numerator = 2^(6x) - 2^(5x) - 2^(4x) + 2^(2x) + 2^x - 1. Using 2^(ax) = 1 + ax·ln2 + (ax·ln2)²/2 + (ax·ln2)³/6 + ...:</p><p>Collect constant terms: 1 - 1 - 1 + 1 + 1 - 1 = 0</p><p>Collect x·ln2 terms: 6 - 5 - 4 + 2 + 1 = 0</p><p>Collect (x·ln2)²/2 terms: (36 - 25 - 16 + 4 + 1)·(ln2)²/2 = 0</p><p>Collect (x·ln2)³/6 terms: (216 - 125 - 64 + 8 + 1)·(ln2)³/6 = 36·(ln2)³/6 = 6(ln2)³</p><p>∴ Numerator ≈ 6(ln2)³·x³ + O(x⁴)</p><p><strong>Step 3:</strong> Denominator = [√(15+cosx) - 4]·sinx. Rationalize: √(15+cosx) - 4 = (15+cosx-16)/(√(15+cosx)+4) = (cosx-1)/(√(15+cosx)+4) ≈ (cosx-1)/8</p><p>Since cosx ≈ 1 - x²/2: cosx - 1 ≈ -x²/2</p><p>∴ Denominator ≈ (-x²/2)/8 · sinx ≈ (-x²/16)·x = -x³/16</p><p><strong>Step 4:</strong> Limit = (6(ln2)³·x³)/(-x³/16) = 6(ln2)³ · (-16) = -96(ln2)³</p><p>∴ Answer: B</p>
Correct Answer: B