lim(x->无穷) x[ln(x+1)-lnx]/2
=lim(x->无穷) x[ln[x(1+1/x)-lnx]/2
=lim(x->无穷) 0.5[ln(1+1/x)]/(1/x) //: 用洛必达法则
=lim(x->无穷) 0.5[-(1/x^2)/(1+1/x) / (-1/x^2)
=lim(x->无穷) 0.5/(1+1/x)
=1/2
极限,lim x趋于无穷 x[ln(x+1)-lnx]/2,怎样得出结果1/2
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