$\lim_{x \to 0} \frac{\ln(1 + x) \cdot \ln(1 - x)}{x^2}$
Step-by-Step Solution
Key Concept: $\ln(1 + x) \approx x - x^2/2$ and $\ln(1 - x) \approx -x - x^2/2$. Product $\approx (x)(-x) = -x^2$ (leading term). Dividing by $x^2$: limit $= -1$.
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Correct Answer: -1