The limit of $((1+x)^{1/x} - e+ ex/2)/x^2$ as $x\to 0$

Probably you have a typo. Notice that by the Taylor series we have

$$(1+x)^{\color{red}{\frac1x}}=e-\frac{ex}2+\frac{11e x^2}{24}+\mathcal O(x^3)$$

We have $$\begin{aligned}F(x)\,&= (1 + x)^{1/x} - e + \frac{ex}{2}\\ &= \exp\left(\frac{\log(1 + x)}{x}\right) - \exp\left(1 + \log\left(1 - \frac{x}{2}\right)\right)\\ &= \exp\left\{1 + \log\left(1 - \frac{x}{2}\right)\right\}\left\{\exp\left(\frac{\log(1 + x)}{x} - 1 - \log\left(1 - \frac{x}{2}\right)\right) - 1\right\}\\ &= e^{u}(e^{v} - 1)\end{aligned}$$ where $$u = 1 + \log\left(1 - \frac{x}{2}\right), v = \frac{\log(1 + x)}{x} - 1 - \log\left(1 - \frac{x}{2}\right)$$ and $u \to 1, v \to 0$ as $x \to 0$. Thus we have $$\begin{aligned}L \,&= \lim_{x \to 0}\frac{F(x)}{x^{2}}\\ &= \lim_{x \to 0}\frac{e^{u}(e^{v} - 1)}{v}\cdot\frac{v}{x^{2}}\\ &= e\cdot\lim_{x \to 0}\frac{1}{x^{2}}\left\{\frac{\log(1 + x)}{x} - 1 - \log\left(1 - \frac{x}{2}\right)\right\}\\ &= e\cdot\lim_{x \to 0}\frac{1}{x^{2}}\left\{\left(1 - \frac{x}{2} + \frac{x^{2}}{3} - \cdots\right) - 1 + \left(\frac{x}{2} + \frac{x^{2}}{8} + \frac{x^{3}}{24} + \cdots\right)\right\}\\ &= e\left(\frac{1}{3} + \frac{1}{8}\right) = \frac{11e}{24}\end{aligned}$$ We can also use LHR instead of series expansion for logs and this will require two applications of LHR to get the following expression $$\begin{aligned}L\, &= \frac{e}{6}\cdot\lim_{x \to 0}\frac{1}{x}\left(-\frac{1}{(1 + x)^{2}} + \frac{4 - x}{(2 - x)^{2}}\right)\\ &= \frac{e}{6}\cdot\lim_{x \to 0}\frac{1}{x}\frac{(4 - x)(1 + x)^{2} - (2 - x)^{2}}{(1 + x)^{2}(2 - x)^{2}}\\ &= \frac{e}{6}\cdot\lim_{x \to 0}\frac{1}{x}\frac{x(11 + x - x^{2})}{(1 + x)^{2}(2 - x)^{2}}\\ &= \frac{11e}{24}\end{aligned}$$