Question

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

Answer 1

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

Answer 2

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