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Let f be a twice-differentiable function with f''(0) = 2. Evaluate lim(x->0) [2f(x) - 3f(2x) + f(4x)] / x².

1. 3
2. 4
3. 6
4. 2

Correct answer: 6

Solution

Using Taylor expansion f(x) = f(0) + f'(0)x + (f''(0)/2)x² +..., the numerator becomes [2-3+1]f(0) + [2-6+4]f'(0)x + [1-6+8](f''(0)/2)x² = 3*(f''(0)/2)*x² = 3x². Dividing by x² gives 3*f''(0)/2... wait: coefficients of x²: 2*(f''(0)/2) - 3*(4f''(0)/2) + (16f''(0)/2) = f''(0) - 6f''(0) + 8f''(0) = 3f''(0) = 6. Limit = 6.

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