<p>\(\lim_{x \to 0} \dfrac{x \tan 2x - 2x \tan x}{(1 - \cos 2x)^2}\) equals</p>
Step-by-Step Solution
Key Concept: Use Taylor expansions for tan x and cos 2x around x = 0 to compare the numerator and denominator orders. The denominator (1 - cos 2x)² behaves like 4x⁴, so you need the numerator's leading term in the same power.
<p><strong>Step 1:</strong> Expand using Taylor series around x = 0.</p><p>tan x = x + x³/3 + 2x⁵/15 + ...</p><p>tan 2x = 2x + 8x³/3 + 64x⁵/15 + ...</p><p><strong>Step 2:</strong> Compute the numerator x tan 2x - 2x tan x:</p><p>x tan 2x = x(2x + 8x³/3 + ...) = 2x² + 8x⁴/3 + ...</p><p>2x tan x = 2x(x + x³/3 + ...) = 2x² + 2x⁴/3 + ...</p><p>Numerator = 2x⁴/3 - 2x⁴/3 + ... = (8x⁴/3 - 2x⁴/3) = 2x⁴ + O(x⁶)</p><p><strong>Step 3:</strong> Compute the denominator (1 - cos 2x)²:</p><p>cos 2x = 1 - 2x² + 2x⁴/3 - ...</p><p>1 - cos 2x = 2x² - 2x⁴/3 + ...</p><p>(1 - cos 2x)² = 4x⁴ - 8x⁶/3 + ...</p><p><strong>Step 4:</strong> Take the ratio:</p><p>lim[x→0] (2x⁴)/(4x⁴) = 1/2</p><p>∴ Answer: <strong>1/2</strong></p>
Correct Answer: C