lim((x→0) (sinx/x)^1/(1-cosx)
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lim((x→0) (sinx/x)^1/(1-cosx)
lim((x→0) (sinx/x)^1/(1-cosx)
lim((x→0) (sinx/x)^1/(1-cosx)
题目应该是x→0+
用自然对数法
lim((x→0+) ln(sinx/x)^1/(1-cosx)
=lim((x→0+) ln(sinx/x)/(1-cosx) (等价无穷小代换)
=lim((x→0+)2[ lnsinx-lnx]/x^2 (洛必达法则)
=lim((x→0+) [ cosx/sinx-1/x]/x (通分)
=lim((x→0+) [ xcosx-sinx]/(x^2sinx) (等价无穷小代换)
=lim((x→0+) [ xcosx-sinx]/x^3 (洛必达法则)
=lim((x→0+) [ cosx-xsinx-cosx]/(3x^2)
=lim((x→0+) [ -xsinx]/(3x^2) (等价无穷小代换)
=-1/3
所以
lim((x→0+) (sinx/x)^1/(1-cosx)
=lim((x→0+) e^[ln(sinx/x)^1/(1-cosx)]
=e^(-1/3)
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