$ \lim_{x\to o} \frac{(1+x)^{\frac1x}-e+\frac{ex}{2}}{ex^2} $ 
$$ \lim_{x\to o} \frac{(1+x)^{\frac1x}-e+\frac{ex}{2}}{ex^2} $$

(can this be duplicate? I think not)
I tried it using many methods
$1.$ Solve this conventionally taking $1^\infty$ form in no luck
$2.$ Did this, expand $ {(1+x)^{\frac1x}}$ using binomial theorem got $\frac13$ then grouped coefficients of $x^0$ and it cancelled with $e$ then took coefficient of $x$ cancelled with $\frac{ex}{2}$ and so on  very messy right ? at last I got $\frac13$ but that's not the expected answer!
I must have went wrong somewhere can anyone help me with this.

We can proceed as follows
\begin{align}
L &= \lim_{x \to 0}\dfrac{(1 + x)^{1/x} - e + \dfrac{ex}{2}}{ex^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + x)}{x}\right) - e + \dfrac{ex}{2}}{ex^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + x)}{x} - 1\right) - 1 + \dfrac{x}{2}}{x^{2}}\text{ (cancelling }e\text{ from Nr and Dr)}\notag\\
&= \lim_{x \to 0}\dfrac{\exp\left(-\dfrac{x}{2} + \dfrac{x^{2}}{3} + o(x^{2})\right) - 1 + \dfrac{x}{2}}{x^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{1 +\left(-\dfrac{x}{2} + \dfrac{x^{2}}{3} + o(x^{2})\right) + \dfrac{1}{2}\left(-\dfrac{x}{2} + \dfrac{x^{2}}{3} + o(x^{2})\right)^{2} + o(x^{2}) - 1 + \dfrac{x}{2}}{x^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{\dfrac{x^{2}}{3} + \dfrac{1}{2}\left(-\dfrac{x}{2} + \dfrac{x^{2}}{3} + o(x^{2})\right)^{2} + o(x^{2})}{x^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{\dfrac{x^{2}}{3} + \dfrac{x^{2}}{8} + o(x^{2})}{x^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{\dfrac{11x^{2}}{24} + o(x^{2})}{x^{2}}\notag\\
&= \frac{11}{24}
\end{align}
Note that using Taylor series seems to be the only option as L'Hospital will only complicate the limit expression.
