$\lim_{x \to 0}\frac{e^{(1+x)^{\frac{1}{x}}}-(1+x)^{\frac{e}{x}}}{x^2}=\quad?\;\;$ (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.
 A: 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*}
