$\lim_{x \to 0} \frac{(1 - \cos 2x)(3 + \cos x)}{x \tan 4x}$
Step-by-Step Solution
Key Concept: $1 - \cos 2x = 2 \sin^2 x$. Denominator: $x \tan 4x \approx 4x^2$. Numerator $\approx 2x^2 \cdot 4 = 8x^2$. So limit $= 8x^2 \cdot 4/(4x^2)...$ regroup: $\frac{2 \sin^2 x}{x^2} \cdot \frac{(3 + \cos x)}{\tan 4x / x} \to 2 \cdot 1 \cdot \frac{4}{4} = 2$.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (3)