作业帮求极限,第一题,
来源：学生作业帮 编辑：拍题作业网作业帮 分类：数学作业 时间：2024/04/20 23:57:38
求极限,第一题,
原式=limx→0 {2ln(2-cosx)-3[(1+sin^2x)^(1/3)-1]}/(x*x)^2 （ln(1+x)~x,替换）
=limx→0 [(2/(2-cosx)*sinx-3*1/3*(1+sin^2x)^(-2/3)*2sinxcosx]/4x^3 (罗比塔法则求导)
=limx→0 2sinx[1/(2-cosx)-cosx(1+sin^2x)^(1/3)/(1+sin^2x)]/4x^3
=limx→0 2x[(1+sin^2x)-cosx(2-cosx)(1+sin^2x)^(1/3)]/4x^3(2-cosx)(1+sin^2x) (sinx~x,替换)
=limx→0 [(1+sin^2x)-cosx(2-cosx)(1+sin^2x)^(1/3)]/2x^2*limx→0 1/(2-cosx)(1+sin^2x)
=limx→0 [(1+sin^2x)-(2cosx-cos^2x)(1+sin^2x)^(1/3)]/2x^2
=limx→0 {2sinxcosx-[(-2sinx+2cosxsinx)(1+sin^2x)^(1/3)+(2cosx-cos^2x)*1/3*(1+sin^2x)^(-2/3)*2sinxcosx]}/4x (罗比塔法则求导)
=limx→0 2sinx{cosx+(1-cosx)(1+sin^2x)^(1/3)-1/3*cosx(2cosx-cos^2x)(1+sin^2x)^(-2/3)}/4x
=limx→0 {cosx+(1-cosx)(1+sin^2x)^(1/3)-1/3*cosx(2cosx-cos^2x)(1+sin^2x)^(-2/3)}/2
=[1+(1-1)(1+0)^(1/3)-1/3*1*(2-1)(1+0)^(-2/3)]/2
=(1-1/3)/2
=1/3.
=limx→0 [(2/(2-cosx)*sinx-3*1/3*(1+sin^2x)^(-2/3)*2sinxcosx]/4x^3 (罗比塔法则求导)
=limx→0 2sinx[1/(2-cosx)-cosx(1+sin^2x)^(1/3)/(1+sin^2x)]/4x^3
=limx→0 2x[(1+sin^2x)-cosx(2-cosx)(1+sin^2x)^(1/3)]/4x^3(2-cosx)(1+sin^2x) (sinx~x,替换)
=limx→0 [(1+sin^2x)-cosx(2-cosx)(1+sin^2x)^(1/3)]/2x^2*limx→0 1/(2-cosx)(1+sin^2x)
=limx→0 [(1+sin^2x)-(2cosx-cos^2x)(1+sin^2x)^(1/3)]/2x^2
=limx→0 {2sinxcosx-[(-2sinx+2cosxsinx)(1+sin^2x)^(1/3)+(2cosx-cos^2x)*1/3*(1+sin^2x)^(-2/3)*2sinxcosx]}/4x (罗比塔法则求导)
=limx→0 2sinx{cosx+(1-cosx)(1+sin^2x)^(1/3)-1/3*cosx(2cosx-cos^2x)(1+sin^2x)^(-2/3)}/4x
=limx→0 {cosx+(1-cosx)(1+sin^2x)^(1/3)-1/3*cosx(2cosx-cos^2x)(1+sin^2x)^(-2/3)}/2
=[1+(1-1)(1+0)^(1/3)-1/3*1*(2-1)(1+0)^(-2/3)]/2
=(1-1/3)/2
=1/3.