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数学
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高数 求极限lim (1+xe^x)^1/sinx x→0
人气:139 ℃ 时间:2019-08-22 12:17:19
解答
原式=lim(1+xe^x)^[(1/xe^x)(xe^x/sinx)] x→0
=e^lim(xe^x/sinx) x→0
=e^lim(xe^x/x) x→0 (sinx与x在x→0时是等价无穷小)
=e^1
=e
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