为什么Lim1/[（e^x) -1] 当 x→1+的极限是∞,x→1-的极限是0,

lim(x->1)1/(e^x-1)=1/(e-1)lim(x->0+)=1/(e^x-1)=+∝e^x>1 x->0+e^x-1->1+lim(x->0-)=1/(e^x-1)=-∝e^x0- e^x-1->1-lim(x->+∝)=1/(e^x-1)=0x->+∝,e^x-1->+∝lim(x->-∝)=1/(e^x-...图片上的例子请帮忙分析，谢谢lime^(1/(x-1))x->1+x>1 x-1->0+ 1/(x-1)->+∝e^(1/(x-1))->+∝x->1- x<1 x-1->0-1/(x-1)->-∝,1/(1-x)->+∝e^(1/(x-1))=1/e^(1/(1-x))->0(x^2-1)/(x-1)=(x+1)lim(x->1+) ( x+1)=lim(x->1-)(x+1)=2lim(x->1+)[(x^2-1)/(x-1)] *e^(1/(x-1))=2*+∝=+∝lim(x->1-)[(x^2-1)/(x-1)] *e^(1/(x-1))=2*0=0lim(x->1+) ≠lim(x->1-) 极限不存在