$ \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.

$(1+x)^{\tfrac 1x}=\mathrm e^{\tfrac{\ln(1+x)}x}$, and $\,
\dfrac{\ln(1+x)}x =1-\dfrac x2+\dfrac{x^2}3+o(x^2)$, hence
\begin{align*}
(1+x)^{\tfrac 1x}-\mathrm e+\frac{\mathrm ex}2&=\mathrm e\Bigl(\mathrm e^{-\tfrac x2+\tfrac{x^2}3+o(x^2)}\Bigr)\\
&=\mathrm e\frac{11x^2}{24}+o(x^2)
\end{align*}
Thus
$$\lim_{x\to o} \frac{(1+x)^{\tfrac1x}-e+\dfrac{ex}{2}}{ex^2}=\lim_{x\to o}\frac{11}{24}+o(1)= \frac{11}{24}.$$
