求极限lim(x->0)(x+e^x)^2/x

来源：学生作业帮编辑：搜搜考试网作业帮分类：数学作业时间：2024/05/04 11:05:33

∵lim(x->0)[ln(x+e^x)/x]=lim(x->0)[(1+e^x)/(x+e^x)] (0/0型极限,应用罗比达法则)
=(1+1)/(0+1)
=2
∴lim(x->0)[(x+e^x)^(2/x)]=lim(x->0){e^[(2/x)ln(x+e^x)]}
=e^{2*lim(x->0)[ln(x+e^x)/x]}
=e^(2*2)
=e^4.

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