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!
人气:489 ℃ 时间:2019-10-23 07:18:38
解答
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...
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