<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 substitution 1 - cos 2x = 2sin²x and the standard limits sin(ax)/ax → a as x → 0, recognizing that tan 4x/x → 4 as x → 0.
<p><strong>Step 1:</strong> Rewrite numerator using 1 - cos 2x = 2sin²x:</p><p>$$\lim_{x \to 0} \frac{2\sin^2 x(3 + \cos x)}{x \tan 4x}$$</p><p><strong>Step 2:</strong> Separate and apply standard limits. Rewrite as:</p><p>$$\lim_{x \to 0} \frac{2\sin^2 x}{x^2} \cdot \frac{x}{\tan 4x} \cdot (3 + \cos x)$$</p><p><strong>Step 3:</strong> Apply standard limits:</p><p>• $\lim_{x \to 0} \frac{\sin x}{x} = 1$, so $\lim_{x \to 0} \frac{\sin^2 x}{x^2} = 1$</p><p>• $\lim_{x \to 0} \frac{\tan 4x}{x} = 4$, so $\lim_{x \to 0} \frac{x}{\tan 4x} = \frac{1}{4}$</p><p>• $\lim_{x \to 0} (3 + \cos x) = 4$</p><p><strong>Step 4:</strong> Combine results:</p><p>$$= 2 \cdot 1 \cdot \frac{1}{4} \cdot 4 = 2$$</p><p>∴ Answer: C (which equals 2)</p>
Correct Answer: C