problem
\[\lim _{x \rightarrow 0} \frac{\mathrm{e}^{(1+x)^{\frac{1}{x}}}-(1+x)^{\frac{e}{x}}}{x^{2}} \]
solution
\[ \begin{aligned} &\text { 解: } \lim _{x \rightarrow 0} \frac{\mathrm{e}^{(1+x)^{\frac{1}{x}}}-(1+x)^{\frac{e}{x}}}{x^{2}}=\lim _{x \rightarrow 0} \frac{\mathrm{e}^{(1+x)^{\frac{1}{x}}}-\mathrm{e}^{\frac{\operatorname{eln}(1+x)}{x}}}{x^{2}} \\ &=\lim _{x \rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-\frac{\mathrm{eln}(1+x)}{x}}{x^{2}} \cdot \mathrm{e}^{\xi}, \text { 其中 } \xi \text { 在 }(1+x)^{\frac{1}{x}} \text { 和 } \frac{\mathrm{eln}(1+x)}{x} \text { 之间 } \Rightarrow \xi \rightarrow \mathrm{e} \\ &=\mathrm{e}^{\mathrm{e}} \lim _{x \rightarrow 0} \frac{\mathrm{e}^{\frac{\ln (1+x)}{x}}-\frac{\mathrm{eln}(1+x)}{x}}{x^{2}}=\mathrm{e}^{\mathrm{e}+1} \lim _{x \rightarrow 0} \frac{\mathrm{e}^{\frac{\ln (1+x)-x}{x}}-\frac{\ln (1+x)-x}{x}-1}{x^{2}} \\ &u=\frac{\ln (1+x)-x}{x} \sim-\frac{1}{2} x, \mathrm{e}^{u}-u-1 \sim \frac{1}{2} u^{2} \sim \frac{1}{8} x^{2} \end{aligned}\]
\[\lim _{x \rightarrow 0} \frac{\mathrm{e}^{(1+x)^{\frac{1}{x}}}-(1+x)^{\frac{e}{x}}}{x^{2}}=\frac{\mathrm{e}^{\mathrm{e}+1}}{8} \]