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Evaluate the limit: lim(x→0) [x * e^(sin x) - e^x * arcsin(sin x)] / [sin²(x) - x * sin(x)]

1. -2
2. -1
3. 1
4. 2

Correct answer: 1

Solution

For x near 0, arcsin(sin x) = x (since |x| < pi/2). So the expression becomes [x*e^(sinx) - x*e^x] / [sin²(x) - x*sin(x)] = x[e^(sinx) - e^x] / [sinx(sinx - x)]. Now sinx ≈ x - x³/6 +..., so sinx - x ≈ -x³/6. e^(sinx) - e^x: let sinx = x - x³/6. e^(sinx) = e^(x - x³/6) = e^x * e^(-x³/6) ≈ e^x*(1 - x³/6). So e^(sinx)-e^x ≈ -e^x * x³/6 ≈ -x³/6 (near x=0 where e^x→1). Numerator: x*(-x³/6) = -x⁴/6. Denominator: sinx*(sinx-x) ≈ x*(-x³/6) = -x⁴/6. Limit = (-x⁴/6)/(-x⁴/6) = 1.

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