题目

$$
\lim_{x\to0}\dfrac{e^{(1+x)^{\frac{1}{x}}} - (1 + x)^{\frac{e}{x}}}{x^2}
$$

解答

$$
\begin{aligned}
&
\lim_{x\to0}\dfrac{e^{(1+x)^{\frac{1}{x}}} - (1 + x)^{\frac{e}{x}}}{x^2}
\\
=&
\lim_{x\to0}\dfrac{e^{(1+x)^{\frac{1}{x}}} - e^{\frac{e\ln(1 + x)}{x}}}{x^2}
\\
=&
e^e \cdot \lim_{x\to0}\dfrac{e^{e^{\frac{\ln(1 + x)}{x}} - \frac{e\ln(1 + x)}{x}} - 1}{x^2}
\\
=&
e^{e + 1} \cdot \lim_{x\to0}\dfrac{e^{\frac{\ln(1 + x) - x}{x}} - \frac{\ln(1 + x)}{x}}{x^2}
\\
=&
e^{e + 1} \cdot \lim_{x\to0}\dfrac{e^{\frac{\ln(1 + x) - x}{x}} - 1 - \frac{\ln(1 + x) - x}{x}}{x^2}
\\
\end{aligned}
$$

$$
e^{\frac{\ln(1 + x) - x}{x}} - 1 \sim \frac{\ln(1 + x) - x}{x} \sim -\dfrac{1}{2}x
$$

$$
e^x - 1 - x \sim \frac{1}{2} x^2
$$

$$
e^{\frac{\ln(1 + x) - x}{x}} - 1 - \frac{\ln(1 + x) - x}{x} \sim \frac{1}{2} \cdot \frac{1}{4} \cdot x^2 = \dfrac{1}{8}x^2
$$

故原式 $ = e^{e + 1} \cdot \lim\limits_{x\to0}\dfrac{\dfrac{1}{8}x^2}{x^2} = \dfrac{1}{8}e^{e + 1}$