1、原式=(sinx/cosx)-sinx
=[sinx(1-cosx)]/cosx]
={sinx*2[sin(x/2)]^2}/cosx
當(dāng)x趨于零時(shí),在乘積的情況下,
有sinx~x, cosx~1, sin(x/2)~(x/2)
所以其主部為(x^3)/2
即tanx-sinx～(x^3)/2
2、因?yàn)閤-sinx為奇函數(shù),只考慮x趨于+0的情形
當(dāng)x屬于(0,∏/2)
有 x-sinx≤tgx-sinx~(x^3)/2
x-sinx≥2sin(x/2)-sinx
=2sin(x/2)*[1-cos(x/2)]
=4sin(x/2)[sin(x/4)]^2~(x^3)/8
即x-sinx～k(x^3)
即lim(x-sinx)/[k*(x^3)]
用洛必達(dá)法則知,當(dāng)x趨于零時(shí),
lim(x-sinx)/[k*(x^3)]=lim(1-cosx)/[3k*(x^2)]
=limsinx/6kx
=lim(1/6k)=1
所以k=1/6
所以x-sinx~(x^3)/6