Question

速求极限问题！！

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Answer 1

1. = sin2x/[(x-pi/2)cos2x] 令y=x-pi/2->0 x=y+pi/2 =sin(2y+pi)/[ycos(2y+pi)] =-sin2y/[y(-cos2y)] =[sin2y/(2y)]*(2/cos2y) 两个因子分别取极限 =1*(2/1) =2 2. =[sin5x/(5x)]*cos2x*[5x/(3x^2+5x)] =[sin5x/(5x)]*cos2x*[5/(5+3x)] 每个因子分别求极限 =1*1*5/5 =1 3. 令 y=x-2->0 x^2-4=(x-2)(x+2)=(x-2)(x-2+4)=y(y+4) 原式=siny/[y(y+4)] =(siny/y)[1/(y+4)] 因子分别取极限 =1*1/4 =1/4

Answer 2

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