如题x趋于 0.求tanx－sinx关于x的阶数，求具体解答讲解

lim(tanx-sinx)/(sinx)^3\u5f53x\u8d8b\u4e8e0\u65f6\u7684\u7ed3\u679c\u600e\u4e48\u6c42\uff1f\u5e2e\u5fd9\u89e3\u91ca\u4e00\u4e0b\uff0c\u8c22\u8c22

lim[ (1/cosx-1)sinx]/sin^3(x)
=lim[(1-cosx)/cosx]/sin^2(x)
=lim[x^2/2cosx]/sin^2(x)
=1/2
\u8fd9\u91cc\u7528\u5230\u4e86x~sinx 1-cosx~x^2/2
\u4eb2\uff0c\u5982\u679c\u6211\u7684\u56de\u7b54\u5bf9\u60a8\u6709\u5e2e\u52a9\uff0c\u8bf7\u8d50\u4e2a\u597d\u8bc4\u5427\u3002\u8c22\u8c22\uff01

limx\u8d8b\u5411\u4e8e0\u65f6\uff0c(e^tanx-e^x)/(x-sinx)

=lim\uff08x->0\uff09e^x(e^(tanx-x)-1)/(x-sinx)
=lim\uff08x->0\uff09(e^(tanx-x)-1)/(x-sinx)
=lim\uff08x->0\uff09(tanx-x)/(x-sinx)

=lim\uff08x->0\uff09(sec²x-1)/(1-cosx)
=lim\uff08x->0\uff09(tan²x)/(1-cosx)

=lim\uff08x->0\uff09(x²)/(x²/2)
=2

tanx－sinx=tanx(1-cosx)=tanx*sin²x/2,
tanx~x,sinx~x
所以是1+2=3阶

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