极限x(e^x+1)-2e^x+2/x-sinx的值是多少,x趋向0,

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用罗比达法则
原式=lim(x→0)(e^x+1+x*e^x-2e^x)/(1-cosx)
=lim(x→0)(xe^x-e^x+1)/(1-cosx)
=lim(x→0)(e^x+x*e^x-e^x)/sinx
=lim(x→0)xe^x/sinx
=lim(x→0)(e^x+xe^x)/cosx
=(e^0+0*e^0)/cos0
=1