设斜渐近线方程：y=kx+b
则k=lim【x→+∞】[(1+x)e^(1-1/x)]/x
=lim【x→+∞】[1/x+1]e^(1-1/x)
=(0+1)e^(1-0)
=e
故b=lim【x→+∞】[(1+x)e^(1-1/x)]-ex
=lim【x→+∞】[e^(1-1/x)+xe^(1-1/x)-ex]
=e
故斜渐近线为：y=ex+e