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lim【x→0+】(1-cosx)^(1/lnx)
人气:313 ℃ 时间:2020-06-25 13:30:10
解答
y=(1-cosx)^(1/lnx)lny=(1/lnx)ln(1-cosx)=(x²/2)/lnx =x²/(2lnx)lim【x→0+】lny=lim【x→0+】x²/(2lnx)=lim【x→0+】(2x)/(2/x)=lim【x→0+】x²=0故lim【x→0+】=1
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