<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] In response to reading some of the comments, I just want to say that the L'hospital calculation is not horrible, if one works a little strategically. Of course, this is not entriely tedium free... 

Let $$g(x) = \frac{\log(1+x)}{x}.$$ We have $g \to 1,$ and $g' \to -1/2$ as $x \to 0.$ (you'll have to show this for $g'$ but this is not hard).
Now let the limit in the question be written as $$L = \lim \frac{f_1(x) - f_2(x)}{x^2}, $$ where
\begin{align} 
f_1(x) &:= e^{e^{g(x)}} \\ f_2(x) &:= e^{eg(x)} \end{align}
(I'll now start dropping the $(x)$ in $g(x)$ for legibility.)
By the chain rule, \begin{align} f_1' &= \left(e^{g(x)}\right)' e^{e^{g}} = g' e^{g} e^{e^{g}} \\ f_2' &= eg' e^{eg}\end{align}
So, by L'Hospital, $$ L = \lim \frac{g'e^g \left(e^{e^g} -e e^{(e-1)g }\right) }{2x}$$
But, $g'e^{g} \to -e/2$, so $$ L = -\frac{e}{4} \lim \frac{e^{e^g} - e^{(e-1)g + 1} }{x} $$ (assuming of course that the latter exists - this should be easy to show).
The point of doing this is get back to something that looks (almost) like $f_1 - f_2$ on the numerator. In particular, I don't want the second derivative of $f_1$ to enter my expressions - that's too much work.
Applying L'Hospital to this again, we get $$ -\frac{4L}{e} = \lim g'e^g e^{e^g} - (e-1)g' e^{(e-1)g + 1} = -\frac{1}{2}e\cdot e^e + \frac{(e-1)}{2}e^e = -\frac{1}{2}e^e,$$ where I've simply evaluated using $g \to 1, g' \to -1/2,$ giving again $$ L = e \cdot e^e/8.$$
</s>