1)x=1。因为x/(x-1)=1+1/(x-1)
当x→1-时,x/(x-1)→-∞
e^[x/(x-1)]→0,y→1
当x→1+时,x/(x-1)→+∞
e^[x/(x-1)]→+∞,y→0
故x=1为跳跃间断点
2)令1-e^[x/(x-1)]=0,得x=0
x→0-时,y→+∞
x→0+时,y→-∞
故x=0为无穷间断点
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