Correct answer: (B) Converges to √2 whenever x₀>√(2/3).
Explanation
This is Newton iteration for x²−2=0. For x₀>0, AM–GM gives x₁≥√2; thereafter √2≤x_(n+1)≤x_n, with limit L=L/2+1/L giving L=√2. Negative initial values remain negative and converge to −√2, refuting A and C. B supplies a positive starting value and is true. **B**.
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