<s>[INST] Evaluate $\lim\limits_{x \to 0}\frac{e^{(1+x)^{\frac{1}{x}}}-(1+x)^{\frac{e}{x}}}{x^2}$ Solution
Expanding $(1+x)^{\frac{1}{x}}$ at $x=0$ by Taylor's Formula，we obtain
\begin{align*}
(1+x)^{\frac{1}{x}}&=\exp\left[\frac{\ln(1+x)}{x}\right]=\exp \left(\frac{x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots }{x}\right)\\
&=\exp \left(1-\frac{x}{2}+\frac{x^2}{3}+\cdots \right)=e\cdot\exp \left(-\frac{x}{2}+\frac{x^2}{3}+\cdots \right) \\
&=e\left[1+\left(-\dfrac{x}{2}+\dfrac{x^2}{3}+\cdots \right)+\frac{1}{2!}\left(-\dfrac{x}{2}+\dfrac{x^2}{3}+\cdots \right)^2+\cdots\right]\\
&=e\left(1-\frac{x}{2}+\frac{11}{24}x^2+\cdots\right)\\
&=e-\frac{ex}{2}+\frac{11}{24}ex^2+\cdots
\end{align*}
Likewise, expanding $e^{(1+x)^{\frac{1}{x}}}$ at $x=0$, we obtain
\begin{align*}
e^{(1+x)^{\frac{1}{x}}}&=(e^e)^{1-\frac{x}{2}+\frac{11}{24}x^2-\cdots}=e^e\cdot (e^e)^{-\frac{x}{2}+\frac{11}{24}x^2+\cdots}\\
&=e^e\cdot\left[1+\left(-\frac{x}{2}+\frac{11}{24}x^2+\cdots\right)\ln e^e+\frac{1}{2!}\left(-\frac{x}{2}+\frac{11}{24}x^2+\cdots\right)^2\ln^2 e^e+\cdots\right]\\
&=e^e \cdot\left[1-\frac{ex}{2}+\frac{1}{24}(11e+3e^2)x^2+\cdots\right]
\end{align*}
Expanding $(1+x)^{\frac{e}{x}}$ at $x=0$, it follows that
\begin{align*}
(1+x)^{\frac{e}{x}}&=\exp\left[\frac{e\ln(1+x)}{x}\right]=\exp \left(e\cdot\frac{x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots }{x}\right)\\
&=\exp \left(e-\frac{ex}{2}+\frac{ex^2}{3}+\cdots \right)=e^e\cdot\exp \left(-\frac{ex}{2}+\frac{ex^2}{3}+\cdots \right) \\
&=e^e\left[1+\left(-\frac{ex}{2}+\frac{ex^2}{3}+\cdots \right)+\frac{1}{2!}\left(-\frac{ex}{2}+\frac{ex^2}{3}+\cdots \right)^2+\cdots\right]\\
&=e^e\left[1-\frac{ex}{2}+\frac{1}{24}e(8+3e)x^2+\cdots\right]
\end{align*}
Therefore
\begin{align*}
&\lim_{x \to 0}\frac{e^{(1+x)^{\frac{1}{x}}}-(1+x)^{\frac{e}{x}}}{x^2}\\
=&\lim_{x \to 0}\frac{e^e\cdot\left[1-\dfrac{ex}{2}+\dfrac{1}{24}e(11+3e)x^2+\cdots\right]-e^e\cdot\left[1-\dfrac{ex}{2}+\dfrac{1}{24}e(8+3e)x^2+\cdots\right]}{x^2}\\
=&e^e\cdot\frac{1}{8}e\\
=&\frac{1}{8}e^{e+1}
\end{align*}
Please check. Is there any more simpler solution?
 [/INST] Looks good, and thumbs up for taking the time to provide your own work!

$$\lim\limits_{x \to 0}\frac{e^{(1+x)^{1/x}}-(1+x)^{e/x}}{x^2}$$

Not sure if this is any better, but taking $t=\left(1+x\right)^{1/x}$ turns the numerator into $e^t-t^e$ with $t \to e$ when $x \to 0$. Its first non-zero term in the series expansion (around $t=e$) is:
$$\frac{1}{2} e^{e - 1} \left(t - e\right)^2 \tag{$\star$}$$
Now borrowing your expansion of $t=\left(1+x\right)^{1/x}$ around $x=0$:

$$(1+x)^{\frac{1}{x}}=\color{blue}{e-\frac{ex}{2}}+\frac{11}{24}ex^2+\ldots$$

And plugging this only up to order $1$ (blue) into $(\star)$ gives:
$$\frac{1}{2} e^{e - 1} \left(\color{blue}{e-\frac{ex}{2}} - e\right)^2 =\frac{1}{8} e^{e + 1} x^2$$

For $(\star)$, with $f(t)=e^t-t^e$ you have:


*

*$f'(t)=e^t-et^{e-1} \implies f'(e)=0$ 

*$f''(t)=e^t-e(e-1)t^{e-2} \implies f''(e)=e^{e-1}$
So:
$$\begin{align}f(t) & = f(e)+f'(e)(t-e)+\frac{1}{2}f''(e)(t-e)^2 + \ldots \\
 &  = 0 + 0 + \frac{1}{2} e^{e - 1} \left(t - e\right)^2 + \ldots\end{align}$$
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