limx趋向0+[x^ln(1+x)]
=limx趋向0+[e^(xln(1+x))]
=e^limx趋向0+[(xln(1+x))]
limx趋向0+(xln(1+x))
=0
所以
limx趋向0+[x^ln(1+x)]=e^0=1