$$\lim_{x\rightarrow 0}\frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}=\,?$$
the given options are
$a)\frac{24e}{11}$
$b)\frac{11e}{24}$
$c)\frac{e}{11}$
$d)\frac{e}{24}$
may be the question would be $$\lim_{x\rightarrow 0}\frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}=\,?$$
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 $$\begin{aligned}L\, &= \frac{e}{6}\cdot\lim_{x \to 0}\frac{1}{x}\left(-\frac{1}{(1 + x)^{2}} + \frac{4 - x}{(2 - x)^{2}}\right)\\ &= \frac{e}{6}\cdot\lim_{x \to 0}\frac{1}{x}\frac{(4 - x)(1 + x)^{2} - (2 - x)^{2}}{(1 + x)^{2}(2 - x)^{2}}\\ &= \frac{e}{6}\cdot\lim_{x \to 0}\frac{1}{x}\frac{x(11 + x - x^{2})}{(1 + x)^{2}(2 - x)^{2}}\\ &= \frac{11e}{24}\end{aligned}$$
We will use three limits derived using L'Hospital:
$$
\lim_{x\to0}\frac{1-e^x}{x}=-1\tag1
$$
and
$$
\lim_{x\to0}\frac{1+x-e^x}{x^2}=-\frac12\tag2
$$
and
$$
\lim_{x\to0}\frac{1+x+\frac12x^2-e^x}{x^3}=-\frac16\tag3
$$
Now
$$
\begin{align}
&\frac{(1+x)^{1/x}-e+\frac{ex}2}{x^2}\\[9pt]
&=\frac{e\left(1+\frac{1+x-e^x}{e^x}\right)^{1/x}-e+\frac{ex}2}{x^2}\tag{4a}\\[6pt]
&=\frac{e}{x^2}\left(\color{#C00}{1}\color{#090}{{+}\frac{1{+}x{-}e^x}{e^x}\frac{\frac1x}1}\color{#00F}{{+}\left(\frac{1{+}x{-}e^x}{e^x}\right)^2\frac{\frac1x\left(\frac1x{-}1\right)}2}{+}O\!\left(x^3\right)\color{#C00}{{-}1}\color{#090}{{+}\frac{x}2}\right)\tag{4b}\\
&=\frac{e}{x^2}\left(\color{#090}{\frac{1+x-e^x+\frac12x^2e^x}{xe^x}}\color{#00F}{+\left(\frac{1+x-e^x}{e^x}\right)^2\frac{\frac1x\left(\frac1x-1\right)}2}+O\!\left(x^3\right)\right)\tag{4c}\\
&=\frac{e}{x^2}\left(\color{#090}{\frac{1{+}x{+}\frac12x^2{-}e^x{+}\frac12x^2\left(e^x{-}1\right)}{xe^x}}\color{#00F}{+\left(\frac{1{+}x{-}e^x}{e^x}\right)^2\frac{\frac1x\left(\frac1x{-}1\right)}2}+O\!\left(x^3\right)\right)\tag{4d}\\
&=e\left(\color{#090}{\frac{1+x+\frac12x^2-e^x}{x^3e^x}+\frac{e^x-1}{2xe^x}}\color{#00F}{+\left(\frac{1+x-e^x}{x^2e^x}\right)^2\frac{1-x}2}+O(x)\right)\tag{4e}\\[3pt]
&\!\!\overset{x\to0}=e\left(\color{#090}{-\frac16+\frac12}\color{#00F}{+\frac18}\right)\tag{4f}\\[6pt]
&=\frac{11}{24}e\tag{4g}
\end{align}
$$
Explanation:
$\text{(4a)}$: $\frac{1+x}{e^x}=1+\frac{1+x-e^x}{e^x}$
$\text{(4b)}$: factor $\frac{e}{x^2}$ out and use the first $3$ terms of the Binomial Theorem
$\text{(4c)}$: cancel the red terms and combine the green terms
$\text{(4d)}$: add and subtract $\frac12x^2$ to the green numerator
$\text{(4e)}$: distribute $\frac1{x^2}$
$\text{(4f)}$: evaluate the limits using $(1)$, $(2)$, and $(3)$
$\text{(4g)}$: simplify
LH is not applicable as $$\lim_{x\rightarrow 0}\frac{(1+x)^{1/2}-e+\frac{ex}{2}}{x^2}$$
is not of the indeterminate form $\dfrac00$ or $\dfrac\infty\infty$ or of the following form
