lim1/(sinx)^2-(cosx)^2/x^2 x趋向于0

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lim1/(sinx)^2-(cosx)^2/x^2 x趋向于0
lim1/(sinx)^2-(cosx)^2/x^2 x趋向于0

lim1/(sinx)^2-(cosx)^2/x^2 x趋向于0
x趋于0时 (cosx)^2=1
所以 为 lim 1/(sinx)^2-1/x^2=0
lim[1/sinx^2-1/x^2]
=(x^2-sinx^2)/x^2*sinx^2
=(x^2-sinx^2)/x^4;
用洛必达法则求导
=(2x-sin2x)/4x^3
=(2-2cos(2x))/12x^2
=4sin(2x)/24x
=8x/24x
=1/3

=lim[(x²-sin²xcos²x)/(sin²x·x²)]
=lim[(x²- 1/4 sin²2x)/(x^4)]
=lim[x²- 1/8 (1-cos4x)]/(x^4)
=lim[2x- 1/8 (sin4x·4)]/(4x³)

全部展开

=lim[(x²-sin²xcos²x)/(sin²x·x²)]
=lim[(x²- 1/4 sin²2x)/(x^4)]
=lim[x²- 1/8 (1-cos4x)]/(x^4)
=lim[2x- 1/8 (sin4x·4)]/(4x³)
=lim[2x- 1/2 ·sin4x]/(4x³)
=lim[2- 2 cos4x]/(12x²)
=lim[2(1- cos4x)]/(12x²)
=lim[2· 1/2 (4x)²]/(12x²)
=lim 16/12
=4/3

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求lim1-cosx^2/x^2sinx^2的极限 lim1/(sinx)^2-(cosx)^2/x^2 x趋向于0 (cosx+2sinx)(cosx-sinx)+sinx2cosx 化简 (sinx+cosx)/(2sinx-cosx)= y=sinx*sinx+2sinx*cosx -sinx/2*cosx/2 证明:2(cosx-sinx)/1+sinx+cosx=cosx/1+sinx-sinx/1+cosx 证明:【2(cosx-sinx)】/(1+sinx+cosx)=cosx/(1+sinx) -sinx/(1+cosx) 求证cosX/(1+sinx)-sinx/(1+cosx)=2(cosx-sinx)/(1+sinx+cosx) 求证：(1+sinx+cosx)/(1+sinx-cosx)-(1+sinx-cosx)/(1+sinx+cosx)=2/tanx 求证:2(sinx-cosx)/(1+sinx+cosx)=sinx/(1+cosx)-cosx/(1+sinx) 求证：cosx/1+sinx-sinx/1+cosx=2(cosx-sinx)/1+sinx+cosx 求证.[(1+sinx+cosx+2sinx cosx)/(1+sinx+cosx)]=sinx+cosx 证明:2(cosx-sinx)/1+sinx+cosx=cosx/1+sinx-sinx/1+cosx 化简((sinx+cosx -1)(sinx-cosx+1)-2cosx)/sin2x 化简（（sinx+cosx-1)(sinx-cosx+1)-2cosx)/sin2x 化简：(sinx)^2/(sinx-cosx)-(sinx+cosx)/((tanx)^2-1) 利用两个重要极限,计算下列极限lim1-cosx/x^2