Evaluate limx→01−cos⁡xcos⁡2x⋯cos⁡nxx2It should be112n(n+1)(2n+1)but I don't know how to prove that. I am also aware that limθ→01−cos⁡θθ2=12 but I don't know how to use that.

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Evaluate limx→01−cos⁡xcos⁡2x⋯cos⁡nxx2It should be112n(n+1)(2n+1)but I don&#039;t know how to prove that. I am also aware that limθ→01−cos⁡θθ2=12 but I don&#039;t know how to use that.

Tyrn7i · Accepted Answer

Use  1−∏k=1ncos⁡kx=1−∏k=1n(1−2sin2kx2)∼2∑k=1n(kx2)2=x22∑k=1nk2

Jason Duke · Accepted Answer

Split the numerator like  1−cos⁡x+cos⁡x(1−cos⁡2xcos⁡3x⋯cos⁡nx)  and the desired limit is equal to   limx→01−cos⁡xx2+limx→0cos⁡x⋅1−cos⁡2x⋯cos⁡nxx2   which is same as   12+limx→01−cos⁡2x⋯cos⁡nxx2   Applying same technique we see that the above is equal to   12+222+limx→01−cos⁡3x⋯cos⁡nxx2   and continuing in same fashion we see that the desired limit is equal to   12+22+⋯+n22=n(n+1)(2n+1)12

Evaluate \lim_{x \to 0} \frac{1-\cos x\cos 2x\cdots \cos nx}{x^2} It should

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