lim（x→0）（e∧x+x）∧1/x

x→0时lim(e^x+x)^(1/x)=lime^ln(e^x+x)^(1/x)=e^limln(e^x+x)^(1/x),而limln(e^x+x)^(1/x)=lim[ln(e^x+x)]/x,
用洛必塔法则lim [ln(e^x+x)] / x= lim [ln(e^x+x)]' / x'=lim (e^x+1) / (e^x+x)=(1+1) / (1+0)=2,
故原式=e^2