设当x→0时,x-(a+bcosx)sinx为x的5阶无穷小,求a,b.

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问题 设当x→0时,x-(a+bcosx)sinx为x的5阶无穷小,求a,b.

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答案x-(a+bcosx)sinx(x→0)=x-asinx-b/2sin2x=x-a[x-x3/3!+x5/5!+o(x5)]-b/2[2x-(2x)3/3!+(2x)5/5!+o(x5)]=(1-a-b)x+(a/6+2b/3)x3-(a/120+2b/15)x5+o(x5),则1-a-b=0,a/6+2b/3=0,解得a=4/3,b=-1/3.

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