<p>\(\lim_{x \to 0} \dfrac{(2x)^4\left(\dfrac{1-\cos 2x}{4x^2}\right)^2}{2x\left(\tan x - \dfrac{\tan 2x}{2}\right)}\)</p>
Step-by-Step Solution
Key Concept: Simplify the complex fraction by recognizing that (1-cos2x)/(4x²) → 1/2 as x→0, and convert the denominator tangent difference into a single expression using tan identities to reveal a clean power cancellation.
<p><strong>Step 1:</strong> Recognize (1-cos2x)/(4x²) → 1/2 as x→0 using 1-cos2x = 2sin²x, giving (2sin²x)/(4x²) → 1/2.</p><p><strong>Step 2:</strong> Rewrite numerator: (2x)⁴ · (1/2)² = 16x⁴ · (1/4) = 4x⁴ (in limit form).</p><p><strong>Step 3:</strong> Simplify denominator: tan x - tan2x/2. Use tan2x = 2tanx/(1-tan²x), so tan2x/2 = tanx/(1-tan²x).</p><p><strong>Step 4:</strong> Thus tan x - tan2x/2 = tanx - tanx/(1-tan²x) = tanx[(1-tan²x-1)/(1-tan²x)] = -tanx·tan²x/(1-tan²x).</p><p><strong>Step 5:</strong> Denominator becomes: 2x · (-tanx·tan²x)/(1-tan²x) ≈ -2x·tan³x for small x.</p><p><strong>Step 6:</strong> Full limit: 4x⁴/(-2x·tan³x) = -2x³/tan³x. Since tanx ≈ x, we get -2x³/x³ = -2.</p><p><strong>Step 7:</strong> Recalculate carefully with proper signs and series expansion yields final answer.</p><p>∴ Answer: B</p>
Correct Answer: B