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lim(x→0+) lncotx/lnx 求极限 ,用洛比达法则
人气:444 ℃ 时间:2019-08-26 06:52:39
解答
lim(x→0+) lncotx/lnx
=lim(x→0+) (1/cotx)*(-csc^2x)/(1/x)
=-lim(x→0+)x/sinxcosx
=-1=lim(x→0+) (1/cotx)*(-csc^2x)/(1/x)=lim(x→0+) (sinx/cosx)*(-1/sin^2x)/(1/x)=-lim(x→0+)x/sinxcosx=-1
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