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求极限lim(x->0)(x+e^x)^2/x
人气:154 ℃ 时间:2019-12-10 10:47:14
解答
∵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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