令x=1/u,则u→0
原极限=lim[u→0] [(1/u)ln(e+u)-1/u]
=lim[u→0] [ln(e+u)-1]/u
洛必达法则
=lim[u→0] 1/(e+u)
=1/e
求极限,lim[xln(e+1/x)-x],x趋向于无穷大
求极限,lim[xln(e+1/x)-x],x趋向于无穷大
数学人气:269 ℃时间:2020-03-28 17:22:08
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