<p>\(\lim_{x \to 0} \dfrac{(1 - \cos 2x)^2}{2x \tan x - x \tan 2x}\) is</p>
Step-by-Step Solution
Key Concept: Use Taylor series expansions around x=0 for cos(2x) and tan(x) to evaluate the indeterminate form 0/0. The numerator expands as (2x²)² = 4x⁴ + O(x⁶), while the denominator requires careful expansion of tan(x) and tan(2x).
<p><strong>Step 1:</strong> Expand the numerator: 1 - cos(2x) = 1 - (1 - 2x² + 2x⁴/3 + ...) = 2x² - 2x⁴/3 + ...</p><p>Therefore: (1 - cos 2x)² = (2x² - 2x⁴/3 + ...)² = 4x⁴ + O(x⁶)</p><p><strong>Step 2:</strong> Expand the denominator using tan(x) = x + x³/3 + 2x⁵/15 + ... and tan(2x) = 2x + 8x³/3 + ...</p><p>2x·tan(x) = 2x(x + x³/3 + ...) = 2x² + 2x⁴/3 + ...</p><p>x·tan(2x) = x(2x + 8x³/3 + ...) = 2x² + 8x⁴/3 + ...</p><p>Denominator = 2x² + 2x⁴/3 - 2x² - 8x⁴/3 = -2x⁴ + O(x⁶)</p><p><strong>Step 3:</strong> Take the limit: lim(x→0) [4x⁴]/[-2x⁴] = -2</p><p>∴ Answer: C</p>
Correct Answer: C