The value of limx→02(x21−cosxcos2x3cos3x……10cos10x) is ______ .
Answer: 55
Solution
Using mcosmx≈1−m1⋅2(mx)2=1−2mx2 for small x, the product cosxcos2x3cos3x⋯10cos10x≈1−2x2(1+2+⋯+10). So x21−product→21(1+2+⋯+10)=21⋅55, and multiplying by 2 gives 55.