问题描述: limx趋于无穷,{ln(x+根号(x^2+1)-ln(x+根号(x^2-1))}/(e^1/x-1)^2求极限急求.感激不尽啊! 1个回答 分类:数学 2014-10-12 问题解答: 我来补答 用泰勒级数和等价无穷小,令t=1/x,求t->0时候的极限即可,此时分母=e^(t)-1->t分子ln(x+√(x^2+1))-ln(x+√(x^2-1))=lnx+ln(1+√1+(1/x^2))-[lnx+ln(1+√1-(1/x^2))]=ln(1+√1+(1/x^2))-ln(1+√1-(1/x^2))=ln(1+√(1+t^2))-ln(1+√(1-t^2))->ln(1+1+t^2/2+o(t^2))-ln(1+1-t^2/2+o(t^2))=ln2+ln(1+t^2/4+o(t^2))-ln2-ln(1-t^2/4+o(t^2))=ln(1+t^2/4+o(t^2))-ln(1-t^2/4+o(t^2))=(t^2/4+o(t^2))-(-t^2/4+o(t^2))=t^2/2所以原极限=lim(t->0) [(t^2/2)/t]=0 展开全文阅读