Evaluate $\lim_{x\to 0}\frac{e^{(x+1)^{1/x}}-(x+1)^{e/x}}{x^2}$

We have that

$$(x+1)^{\frac{1}{x}}=e^{\frac1x \log (1+x)}=e^{1-\frac12 x+\frac13x^2+o(x^2)}=e \cdot e^{-\frac12 x+\frac13x^2+o(x^2)}=e\left(1-\frac 1 2 x+\frac 1 3 x^2 +\frac12\left(-\frac12 x+\frac13x^2\right)^2+ o(x^2)\right) =e\left(1-\frac 1 2 x+\frac{11}{24}x^2+o(x^2)\right)$$

$$e^{{(x+1)}^{\frac{1}{x}}}=e^{e}e^{-\frac e 2 x+\frac{11e}{24}x^2+o(x^2)}=e^{e}\left(1-\frac e 2 x+\frac{11e}{24}x^2+\frac{e^2}{8}x^2+o(x^2)\right)$$

and

$$(x+1)^{\frac{e}{x}}=e^{\frac e x \log (1+x)}=e^{e-\frac e 2 x+\frac e3x^2+o(x^2)}=e^ee^{-\frac e 2 x+\frac e3x^2+o(x^2)}=e^e\left(1-\frac e 2 x+\frac e3x^2+\frac {e^2}{8}x^2+o(x^2)\right)$$

and thus

$$\dfrac{e^{(x+1)^{\frac{1}{x}}}-(x+1)^{\frac{e}{x}}}{x^2}=\frac{e^{e}\left(1-\frac e 2 x+\frac{11e}{24}x^2+\frac{e^2}{8}x^2+o(x^2)\right)-e^e\left(1-\frac e 2 x+\frac e3x^2+\frac {e^2}{8}x^2+o(x^2)\right)}{x^2}=$$

$$=\frac{\frac{11e^{e+1}}{24}x^2-\frac{e^{e+1}}{3}x^2+o(x^2)}{x^2}=\frac{\frac{e^{e+1}}{8}x^2+o(x^2)}{x^2}=\frac{e^{e+1}}{8}+o(1)\to \frac{e^{e+1}}{8}$$

The numerator is of the form $a-b$ where $a, b$ both tend to $e^e$. It can be rewritten as $$b\cdot\frac{\exp(\log (a/b)) - 1}{\log(a/b)}\cdot \log(a/b) $$ and the fraction in middle tends to $1$ (because $a/b\to 1$) and hence the desired limit is equal to the limit of expression $$e^e\cdot\frac{\log a - \log b} {x^2}$$ or $$e^e\cdot\dfrac{(1+x)^{1/x}-\dfrac{e}{x}\log(1+x)} {x^2}$$ Applying the same process again (numerator in form $a-b$ with $a, b$ tending to same number $e$ here) we can see that the desired limit is equal to the limit of $$e^e\cdot e\cdot\dfrac{\dfrac{\log(1+x)}{x}-1-\log\left(\dfrac{\log(1+x)}{x}\right)} {x^2}\tag{1}$$ This limit can now be evaluated via Taylor series without much effort. We have $$\frac{1}{x}\log(1+x)=1-\frac{x}{2}+\frac{x^2}{3}+o(x^2)$$ and the logarithm of the above series is $$-\frac{x} {2}+\frac{x^2}{3}-\frac{x^2}{8}+o(x^2)$$ Thus the desired limit is $e^{e+1}/8$.

On a second thought the evaluation of limit of expression $(1)$ can be greatly simplified (and one can also use L'Hospital's Rule) after some manipulation. Barring the constant factor $e^{e+1} $ the fraction in $(1)$ can be written as $$\frac{u-\log(1+u)}{u^2}\cdot\left(\frac{u}{x}\right)^2\tag{2}$$ where $$u=\frac{\log(1+x)-x}{x}\to 0$$ as $x\to 0$ and by L'Hospital's Rule the first fraction in $(2)$ tends to $1/2$ and $u/x\to - 1/2$ so that the expression in $(2)$ tends to $1/8$. Thus the complicated limit is handled using just a single application of L'Hospital's Rule. The same conclusion can be reached via Taylor series without any computation of coefficients.

You may approach $$\lim_{x\to 0}\frac{\exp\exp\frac{\log(x+1)}{x}-\exp\left(e\frac{\log(x+1)}{x}\right)}{x^2}$$ by composition of Maclaurin series. We have $$ \frac{\log(1+x)}{x}=1-\frac{x}{2}+\frac{x^2}{3}+O(x^3) $$ $$ \exp\left(\frac{\log(x+1)}{x}\right)=e-\frac{e}{2}x+\frac{11e}{24} x^2+O(x^3) $$ $$ \exp\left(e\frac{\log(x+1)}{x}\right)=e^e-\frac{1}{2}e^{e+1}x+\frac{1}{24}e^{e+1}(8+3e)x^2+O(x^3) $$ $$ \exp\exp\frac{\log(x+1)}{x}=e^e-\frac{1}{2}e^{e+1}x+\frac{1}{24}e^{e+1}(11+3e)x^2+O(x^3) $$ hence the wanted limit equals $\frac{1}{8}e^{e+1}$.