<p>\(\lim_{x \to 0} \dfrac{(1-\cos 2x)(3+\cos x)}{x \tan 4x}\) is equal to</p>
Step-by-Step Solution
Key Concept: Use the small angle approximations: 1 - cos(2x) ≈ 2x² and tan(4x) ≈ 4x as x → 0, then simplify algebraically before substituting limits.
<p><strong>Step 1:</strong> Recognize the 0/0 indeterminate form. Rewrite using standard limits.</p><p><strong>Step 2:</strong> Separate the expression: $\lim_{x \to 0} \dfrac{(1-\cos 2x)}{x \tan 4x} \cdot (3+\cos x)$</p><p><strong>Step 3:</strong> Use $1-\cos 2x = 2\sin^2 x$ and rewrite: $\lim_{x \to 0} \dfrac{2\sin^2 x}{x \tan 4x} \cdot (3+\cos x)$</p><p><strong>Step 4:</strong> Apply standard limits: $\dfrac{\sin x}{x} \to 1$ and $\dfrac{\tan 4x}{x} \to 4$</p><p><strong>Step 5:</strong> Simplify: $\lim_{x \to 0} \dfrac{2\sin^2 x}{x^2} \cdot \dfrac{x}{\tan 4x} \cdot (3+\cos x) = 2 \cdot 1 \cdot \dfrac{1}{4} \cdot (3+1) = \dfrac{2 \cdot 4}{4} = 2$</p><p>∴ Answer: <strong>B (2)</strong></p>
Correct Answer: B