<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] Your result is correct. A simpler method is to decompose this limit into a product of simpler limits:
Let $f(x)=(1+x)^{1/x}$. We have $\lim_{x\rightarrow 0} f(x) = e$. Note that your expression can be written as
$$ \lim_{x\rightarrow 0} \frac{e^{f(x)}- f(x)^e}{x^2} = \lim_{x\rightarrow 0} \frac{e^{f(x)}- f(x)^e}{(f(x)-e)^2}\frac{(f(x)-e)^2}{x^2} = \lim_{y\rightarrow e} \frac{e^y - y^e}{(y-e)^2} \cdot \left(\lim_{x\rightarrow 0} \frac{f(x)-e}{x}\right)^2 $$
From l'Hôpital's rule we have
$$ \lim_{y\rightarrow e} \frac{e^y - y^e}{(y-e)^2} =^H \lim_{y\rightarrow e} \frac{e^y - ey^{e-1}}{2(y-e)} =^H \lim_{y\rightarrow e} \frac{e^y - e(e-1)y^{e-2}}{2} = \frac12 e^{e-1}$$
$$ \lim_{x\rightarrow 0} \frac{f(x)-e}{x} = \lim_{x\rightarrow 0} \frac{e^{\ln(1+x)/x}- e}{\ln(1+x)/x - 1}\frac{\ln(1+x)/x - 1}{x} = \lim_{y\rightarrow 1}\frac{e^y -e}{y-1} \lim_{x\rightarrow 0}\frac{\ln(1+x)-x}{x^2} $$
$$ \lim_{y\rightarrow 1}\frac{e^y -e}{y-1} =^H \lim_{y\rightarrow 1} \frac{e^y}{1} = e $$
$$ \lim_{x\rightarrow 0}\frac{\ln(1+x)-x}{x^2} =^H \lim_{x\rightarrow 0}\frac{\frac{1}{1+x}-1}{2x} = \lim_{x\rightarrow 0}\frac{-1}{2(1+x)} = -\frac12$$
In total
$$ \lim_{x\rightarrow 0} \frac{e^{f(x)}- f(x)^e}{x^2} = \frac12 e^{e-1} \cdot\left(-\frac{e}{2}\right)^2 = \frac{e^{e+1}}{8}$$
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