Question

利用等价无穷小的性质求极限 limx→0(tanx-tanx-sinx)/(arctanx)^3

Answer 1

问：利用等价无穷小的性质求极限 limx→0(tanx-tanx-sinx)/(arctanx)^3

答：亲，您好，原式=limx→0 [tanx-tan(sinx)]/x^3*limx→0 sinx/x*limx→0 x/arctanx=limx→0 [1/cos^2x-cosx/cos^2(sinx)]/3x^2*1*1=limx→0 [cos^2(sinx)-cos^3x]/3x^2*limx→0 1/[cos^2x*cos^2(sinx)]=limx→0 [-2cos(sinx)sin(sinx)cosx+3cos^2xsinx]/6x*1=limx→0 [3sin2x-2sin(2sinx)]/12x*limx→0 cosx=limx→0 [6cos2x-4cos(2sinx)*cosx]/12=(6-4)/12=1/6. 希望我的回答能够帮到您哦。