xin影相随
幼苗
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lim[x->+∞](x/(x-1))^√x
=lim[x->+∞]e^(√xln(x/(x-1)))
考察lim[x->+∞]√xln(x/(x-1))
=lim[x->+∞]ln(x/(x-1))/(1/√x) (0/0型,可以用罗比塔法则)
=lim[x->+∞](ln(x/(x-1)))'/(1/√x)'
=lim[x->+∞]2√x/(x-1)
=0
所以lim[x->+∞](x/(x-1))^√x=e^0=1
1年前
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