[A202211110354](2022, 南开大学)设 \(x_n=\displaystyle\sum_{k=0}^n{\frac{1}{k!}}\)，\(n=1,2,\cdots\)，求极限
\[\lim_{n\rightarrow \infty} \left( \frac{\ln x_n}{\sqrt[n]{\mathrm{e}}-1}-n \right) . \]解. 显然 \(\mathrm{e}=x_n+\dfrac{1}{\left( n+1 \right) !}+\dfrac{\mathrm{e}^{\theta}}{\left( n+2 \right) !}\)，其中\(0<\theta <1\)，从而
\[\begin{aligned} \lim_{n\rightarrow \infty} \left( \frac{\ln x_n}{\mathrm{e}^{\frac{1}{n}}-1}-n \right) &=\lim_{n\rightarrow \infty} \left( \frac{\ln x_n-n\left( \sqrt[n]{\mathrm{e}}-1 \right)}{\frac{1}{n}} \right) \\ &=\lim_{n\rightarrow \infty} \left( \frac{\ln x_n-n\left( \frac{1}{n}+\frac{1}{2n^2}+o\left( \frac{1}{n^2} \right) \right)}{\frac{1}{n}} \right) \\ &=\lim_{n\rightarrow \infty} n\left( \ln x_n-1-\frac{1}{2n} \right) \\ &=\lim_{n\rightarrow \infty} n\ln \frac{x_n}{\mathrm{e}}-\frac{1}{2}. \end{aligned} \]以下考虑\(\displaystyle\lim_{n\rightarrow \infty} n\ln \frac{x_n}{\mathrm{e}}\)，
\[\begin{aligned} \lim_{n\rightarrow \infty} n\ln \frac{x_n}{\mathrm{e}}&=\lim_{n\rightarrow \infty} n\ln \left( 1-\frac{1}{\mathrm{e}}\left( \frac{1}{\left( n+1 \right) !}+\frac{\mathrm{e}^{\theta}}{\left( n+2 \right) !} \right) \right) \\ &=-\frac{1}{\mathrm{e}}\lim_{n\rightarrow \infty} \left( \frac{n}{\left( n+1 \right) !}+\frac{n\mathrm{e}^{\theta}}{\left( n+2 \right) !} \right) \\ &=0. \end{aligned} \]故 \(\displaystyle\lim_{n\rightarrow \infty} \left( \dfrac{\ln x_n}{\sqrt[n]{\mathrm{e}}-1}-n \right) =-\frac{1}{2}.\)
标签：A202211110354,right,frac,lim,left,mathrm,rightarrow From： https://www.cnblogs.com/bairemath/p/16879397.html