Question:

<p><strong>Step 1:</strong> Use the identity 1 - cos(2x) = 2sin²(x):</p><p>$$\lim_{x \to 0} \frac{2\sin^2(x)(3+\cos x)}{x\tan 4x}$$</p><p><strong>Step 2:</strong> Rewrite tan(4x) = sin(4x)/cos(4x) and rearrange:</p><p>$$\lim_{x \to 0} \frac{2\sin^2(x)(3+\cos x) \cos(4x)}{x \sin(4x)}$$</p><p><strong>Step 3:</strong> Separate into standard limit forms:</p><p>$$\lim_{x \to 0} \frac{\sin(x)}{x} \cdot \frac{\sin(x)}{x} \cdot \frac{\sin(4x)}{4x} \cdot \frac{4}{1} \cdot (3+\cos x) \cdot \cos(4x) \cdot \frac{1}{\sin(4x)}$$</p><p><strong>Step 4:</strong> Apply standard limits: $\lim_{u \to 0}\frac{\sin u}{u} = 1$</p><p>$$= 1 \cdot 1 \cdot 1 \cdot 4 \cdot (3+1) \cdot 1 = 4 \cdot 4 = \frac{3}{2}$$</p><p><strong>Correction:</strong> $$= \frac{2 \cdot 1 \cdot 1 \cdot (3+1)}{1 \cdot 4} = \frac{2 \cdot 4}{4} = 2$$</p><p>∴ Answer: C</p>