$$ \begin{align} &\lim_{x\to+\infty}\frac{(1+\frac{1}{x})^{x^2}}{𝕖^x}=\lim_{x\to+\infty}\exp\{x^2\log(1+\frac{1}{x})-x\} \\ &\lim_{x\to+\infty}x^2\log(1+\frac{1}{x})-x=\lim_{x\to+\infty}\frac{\log(1+\frac{1}{x})-\frac{1}{x}}{\frac{1}{x^2}}=\lim_{t\to0+}\frac{\log(1+t)-t}{t^2} \\ &=\lim_{t\to0+}\frac{\frac{1}{1+t}-1}{2t}=-\frac{1}{2} \\ &\lim_{x\to+\infty}\frac{(1+\frac{1}{x})^{x^2}}{𝕖^x}=𝕖^{-\frac{1}{2}}\texttt{，并非1。} \\ &\texttt{同理，}\lim_{x\to+\infty}(1-\frac{1}{x})^{x^2}{𝕖^x}=𝕖^{-\frac{1}{2}}\texttt{。} \end{align} $$