<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] Since both terms, say $A, B$, in numerator tend to same limit $e^e$ we can write $$A-B=B\cdot\frac{\exp (\log A - \log B) - 1}{\log A - \log B} \cdot(\log A - \log B) $$ and thus the numerator can be replaced by $e^e(\log A - \log B) $ or $$e^e\left((1+x)^{1/x}-e\cdot\frac{\log (1+x)}{x}\right)$$ Applying the same technique the above expression can be replaced by $$e^e\cdot e\left(\frac{\log(1+x)}{x}-1-\log\frac{\log(1+x)}{x}\right)$$ or $$e^{e+1}\cdot(u-\log(1+u))$$ where $$u=\frac{\log(1+x)}{x}-1\to 0$$ On the other hand we also know via L'Hospital's Rule or Taylor series that $u/x\to - 1/2$ and therefore the expression $$\frac{u-\log(1+u)}{x^2}=\frac{u-\log(1+u)}{u^2}\cdot\frac{u^2}{x^2}$$ tends to $(1/2)(-1/2)^2=1/8$. The desired limit is thus $e^{e+1}/8$.
In general one should avoid multiplication/division and composition of Taylor series and use the famous ones directly from memory. Often the use of algebraic manipulation combined with standard limits reduces / alleviates the need of any gymnastics with Taylor series. 
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