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If lim(n->infinity) [e*(1 - 1/n)ⁿ - 1] / n^alpha exists and equals l (l not equal to 0), find the value of 12*(l - alpha).

1. 4
2. 3
3. 6
4. 7

Correct answer: 6

Solution

Let's compute the limit. We have (1 - 1/n)ⁿ = e^(n*ln(1-1/n)). Taylor series: ln(1 - 1/n) = -1/n - 1/(2n²) - 1/(3n³) -... So n*ln(1-1/n) = -1 - 1/(2n) - 1/(3n²) -... Therefore (1-1/n)ⁿ = e^(-1) * e^(-1/(2n) + O(1/n²)) = (1/e) * (1 - 1/(2n) + O(1/n²)). Thus e*(1-1/n)ⁿ = 1 - 1/(2n) + O(1/n²). So e*(1-1/n)ⁿ - 1 = -1/(2n) + O(1/n²). For the limit [e*(1-1/n)ⁿ - 1]/n^alpha to be finite and non-zero as n->infinity: we need alpha = -1, and then the limit l = -1/2. Therefore 12*(l - alpha) = 12*(-1/2 - (-1)) = 12*(1/2) = 6.

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