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Define f(x) = x cos(x). Evaluate lim(x->0) { [f(x)/sin(x)] + [f(2x)/sin(2x)] + [f(3x)/sin(3x)] +... + [f(2015x)/sin(2015x)] }, where [.] denotes the greatest integer (floor) function.

1. 2015
2. 4030
3. 2015/2
4. 0

Correct answer: 0

Solution

Each term kx*cos(kx)/sin(kx) approaches 1 from below as x->0 (since cos(kx) < 1 and sin(kx)/(kx) < 1 for small x > 0, their combined effect gives a value slightly less than 1). The floor of any value in (0,1) is 0, so each of the 2015 terms contributes 0, and the total sum is 0.

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