$\lim_{x \to 0}\frac{e^{(1+x)^{\frac{1}{x}}}-(1+x)^{\frac{e}{x}}}{x^2}=\quad?\;\;$ (without using Taylor's Theorem) It is clear that this problem could use Taylor's Theorem to expand the numerator as series, which leads to the solution easily.
But I don't think this method is beautiful enough.
Is there any better method to solve this problem? Any answer would be highly appreciated.

This may not be as beautiful as you were hoping, but here is a solution:
Add and subtract $e^e$ and break the limit into two:
\begin{align*}
\lim_{x\to 0}\frac{e^{(1+x)^{1/x}}-e^e}{x^2}-\frac{(1+x)^{e/x}-e^e}{x^2}&=\lim_{x\to 0}e^e\left[\frac{e^{(1+x)^{1/x}-e}-1}{x^2}-\frac{e^{e/x\ln(1+x)-e}-1}{x^2}\right]
\end{align*}
For the first term, set $\alpha=(1+x)^{1/x}-e$. Note that $\alpha\to 0$ as $x\to 0$. Then we have
\begin{align*}
\lim_{x\to0}\frac{e^{\alpha}-1}{\alpha}\cdot\frac{\alpha}{x^2}=\lim_{x\to0}\frac{\alpha}{x^2}
\end{align*}
For the second term, set $\beta=\frac ex\ln(1+x)-e$. Note that $\beta\to 0$ as $x\to 0$. Then we have
\begin{align*}
\lim_{x\to0}\frac{e^{\beta}-1}{\beta}\cdot\frac{\beta}{x^2}=\lim_{x\to0}\frac{\beta}{x^2}
\end{align*}
Note that I am able to evaluate the first factors of the above limits to $1$, since both terms of the original limit have the "same" factor. Therefore, our original limit becomes
\begin{align*}
\lim_{x\to 0}e^e\left[\frac{\alpha-\beta}{x^2}\right]&=\lim_{x\to 0}e^e\left[\frac{(1+x)^{1/x}-e}{x^2}-\frac{\frac ex\ln(1+x)-e}{x^2}\right]\\
&=\lim_{x\to0}e^e\left[\frac{e^{1/x\ln(1+x)}-e}{x^2}-\frac{\frac ex\ln(1+x)-e}{x^2}\right]\\
&=\lim_{x\to0}e^{e+1}\left[\frac{e^{1/x\ln(1+x)-1}-1}{x^2}-\frac{\frac1x\ln(1+x)-1}{x^2}\right]
\end{align*}
We will only look at the first term. Set $\gamma=\frac1x\ln(1+x)-1$. Then we have
\begin{align*}
\lim_{x\to0}\frac{e^\gamma-1}{\gamma}\cdot\frac{\gamma}{x^2}
\end{align*}
So, our original limit becomes
\begin{align*}
\lim_{x\to0}e^{e+1}\left[\frac{e^\gamma-1}{\gamma}\cdot\frac{\gamma}{x^2}-\frac{\gamma}{x^2}\right]&=\lim_{x\to0}e^{e+1}\cdot\frac{\gamma}{x}\cdot\left(\frac{e^\gamma-1-\gamma}{\gamma x}\right)\\
&=\lim_{x\to0}e^{e+1}\cdot\frac{\ln(1+x)-x}{x^2}\cdot\left(\frac{e^\gamma-1-\gamma}{\gamma x}\right)
\end{align*}
The first factor give us $e^{e+1}$, the middle one can be found using l'Hopitals rule twice, which gives us $-1/2$. The final term evaluates to $-1/4$, but I don't have time to try and find out how at the moment.
Edit: The limit is finished below.
To get the final term, apply l'Hopital's rule twice to get
\begin{align*}
\lim_{x\to 0}\frac{e^\gamma(\gamma'^2+\gamma'')-\gamma''}{\gamma''x+2\gamma'}.
\end{align*}
Now,
\begin{align*}
\gamma'&=\frac{1}{1+x}\cdot\left(\frac{x-\ln(1+x)}{x^2}-\frac{\ln(1+x)}{x}\right)\\
\gamma''&=\frac{1}{(1+x)^2}\cdot\left(\frac{2\ln(1+x)}{x}+\frac{2\ln(1+x)+4x\ln(1+x)-3x^2-2x}{x^3}\right)
\end{align*}
Therefore,
\begin{align*}
\lim_{x\to0}\gamma'&=1\cdot\left(\frac12-1\right)=-\frac12\\
\lim_{x\to0}\gamma''&=1\cdot\left(2-\frac43\right)=\frac23
\end{align*}
Hence,
$$
\lim_{x\to 0}\frac{e^\gamma(\gamma'^2+\gamma'')-\gamma''}{\gamma''x+2\gamma'}=\frac{e^0((-1/2)^2+2/3)-2/3}{2/3(0)+2(-1/2)}=-\frac14.
$$
Putting this all together gives us our final answer:
\begin{align*}
\lim_{x\to0}e^{e+1}\cdot\frac{\ln(1+x)-x}{x^2}\cdot\left(\frac{e^\gamma-1-\gamma}{\gamma x}\right)=\frac{e^{e+1}}8
\end{align*}
