Question

请问这道题求极限用泰勒公式怎么求?

Answer 1

记y=(1+x^2)^(1/2),利用Taylor展开得到y=1+1/2*x^2+o(x^3)

1/ln(x+y)-1/ln(1+x)

=[ln(1+x)-ln(x+y)]/[ln(1+x)ln(x+y)]

再做Taylor展开得到

ln(1+x)=x-1/2*x^2+o(x^3)

ln(x+y)=ln(1+x+1/2*x^2+o(x^3))=x+1/2*x^2+o(x^3)-1/2*[x+1/2*x^2+o(x^3)]^2+o[x+1/2*x^2+o(x^3)]^3=x+o(x^3)

代进去得到

ln(1+x)-ln(x+y)=-1/2*x^2+o(x^3)

ln(1+x)ln(x+y)=x^2+o(x^3)

所以[ln(1+x)-ln(x+y)]/[ln(1+x)ln(x+y)]-> -1/2