求

lim⁡x→0e(1+x)1x−(1+x)exx2\lim_{x \to 0}\frac{{e}^{(1+x)^{\frac{1}{x}}}-(1+x)^{\frac{{e}}{x}}}{x^2}

解答：

记：y=ln(1+x)x有：lim⁡x→0y=1y’=−ln(1+x)x2+1x(1+x)则：lim⁡x→0dydx=lim⁡x→0y′=−12\begin{aligned} 记：&y =\frac{ln{(1+x)}}{x}\\\\ 有：&\lim_{x \to 0} y = 1\\\\ &y’ = -\frac{ln(1+x)}{x^2}+\frac{1}{x(1+x)}\\\\ 则：&\lim_{x \to 0}\frac{dy}{dx}= \lim_{x \to 0} y'=-\frac{1}{2}\\\\ \end{aligned}

所以：原式=lim⁡x→0eey−eeyx2=lim⁡x→0eey+y−eey+12x⋅dydx=lim⁡x→0eey+1−eey+y4x=lim⁡x→0eey+1edydx−eey+y(eydydx+dydx)4=lim⁡x→0eey+1e⋅(−12)−eey+y(ey⋅(−12)+−12)4=lim⁡y→1ee+1⋅e⋅(−12)−ee+1⋅e⋅(−12)−ee+1⋅(−12)4=ee+18\begin{aligned} 所以：原式&=\lim_{x \to 0}\frac{{e}^{{e}^y}-{e}^{{e}y}}{x^2}\\\\ &=\lim_{x \to 0}\frac{{e}^{{e}^y+y}-{e}^{{e}y+1}}{2x} \cdot \frac{dy}{dx}\\\\ &=\lim_{x \to 0}\frac{{e}^{{e}y+1}-{e}^{{e}^y+y}}{4x}\\\\ &=\lim_{x \to 0}\frac{{e}^{{e}y+1}{e}\frac{dy}{dx}-{e}^{{e}^y+y}({e}^{y}\frac{dy}{dx}+\frac{dy}{dx})}{4}\\\\ &=\lim_{x \to 0}\frac{{e}^{{e}y+1}{e}\cdot (\frac{-1}{2})-{e}^{{e}^y+y}({e}^{y}\cdot (\frac{-1}{2})+\frac{-1}{2})}{4}\\\\ & = \lim_{y \to 1}\frac{{e}^{{e}+1} \cdot {e} \cdot (\frac{-1}{2})-{e}^{{e}+1} \cdot {e} \cdot (\frac{-1}{2})-{e}^{{e}+1} \cdot (\frac{-1}{2})}{4}\\\\ &= \frac{{e}^{{e}+1}}{8} \end{aligned}