求极限 lim(x→+∞)[ln(1+x^2)／(2lnx)]^(x^2lnx)

elim

题: 计算\(\;\displaystyle\lim_{x\to\infty}\left(\frac{\ln(1+x^2)}{2\ln x}\right)^{x^2\ln x}\)

elim

\(\frac{\ln(1+x^2)}{2\ln x} = 1+\frac{\ln(1+1/x^2)}{2\ln x}=1+o(1)\)

elim

\(\left(\frac{\ln(1+x^2)}{2\ln x}\right)^{x^2\ln x}=\left(1+\frac{\ln(1+x^{-2})}{2\ln x}\right)^{x^2\ln x}\)
\(=\left(\left(1+\frac{\ln(1+x^{-2})}{2\ln x}\right)^{\tiny\dfrac{2\ln x}{\ln(1+x^{-2})}}\right)^{{\tiny\dfrac{\ln(1+x^{-2})}{2\ln x}}x^2\ln x}=\left(\left(1+\frac{\ln(1+x^{-2})}{2\ln x}\right)^{\tiny\dfrac{2\ln x}{\ln(1+x^{-2})}}\right)^{{\tiny\dfrac{\ln(1+x^{-2})^{x^2}}{2
}}}\overset{x\to\infty}{\longrightarrow} e^{\frac{\ln e}{2}}=\sqrt{e}\)

luyuanhong

楼上 elim 的帖子很好！已收藏。