f(x)’=[x^2*(e^x-1)-2x*(e^x-1-x)]/x^4
=[x^2*e^x-x^2-2x*e^x+2*x+2*x^2]/x^4
=[x^2*e^x-2x*e^x+2*x+x^2]/x^4
x05x05
设g(x)= x^2*e^x-2x*e^x+2*x+x^2
g(x)’=x^2*e^x+e^x*2x-2*e^x-2x*e^x+2+2*x
= x^2*e^x-2*e^x+2+2*x
g(x)’’=x^2*e^x+e^x*2x-2*e^x+2
g(x)’’’= x^2*e^x+2x* e^x+ e^x*2+ e^x*2x-2*e^x
= x^2*e^x+2x* e^x+ e^x*2x
>0
g(x)’’单调递增
g（x）''>=g’’(0)=0
g(x)’单调递增
g（x）'>=g(0)=0
g(x)单调递增
g(x)>=g(0)=0
又f(x)’=g(x)/x^4>0
f(x)单调递增
f（x）min=f(0)=lim （e^x-1）/2x=lim e^x=1/2(洛必达法则)
以上x→0
综上 f(x)>1/2