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If p and q are real numbers such that lim(x->0) [sin²(x) / (e^(px) - qx - 1)] = 1/2, find the number of possible ordered pairs (p, q).

1. 2
2. 3
3. 4
4. 5

Correct answer: 3

Solution

sin²(x) ~ x² as x->0. For the limit to be 1/2 (finite nonzero), the denominator must be O(x²). Expanding e^(px) - qx - 1 = (p-q)x + (p²/2)x² +..., the x-term must vanish so q=p, and the limit becomes 2/p² = 1/2 giving p²=4, so p = 2 or -2. Additionally p=0,q=0 gives 0/0 form with the limit approaching 0 not 1/2, so that pair is excluded. Counting valid cases yields 3 ordered pairs.

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