$ \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.
 A: let $y = \left(1 + \frac1n\right)^n.$ taking logarithm, we get 
$$\begin{align} \ln y &= n\ln\left(1 + \frac 1n  \right)\\
           &=  n \left(\frac 1n - \frac1{2n^2}  +\frac1{3n^3}+\cdots \right)  \\
           & = 1 - \frac1{2n} + \frac{1}{3n^2}+ \cdots        \end{align}$$
therefore $$\begin{align} y &= ee^{-\frac1{2n}} e^{\frac1{3n^2}}\\
&=e\left(1-\frac{1}{2n}+ \frac1{8n^2}+\cdots\right) 
\left(1+\frac{1}{3n^2}+\cdots\right) \\
&=e\left(1- \frac1{2n} +\frac{11}{24n^2} + \cdots\right)
\end{align}$$
from this we get $$n^2\left(y - e + \frac e{2n}\right) = \frac{11e}{24} + \cdots $$ and $$\lim_{x \to 0}\frac{(1+ x)^{\frac1x}- e + \frac{ex}2}{ex} =\frac{11}{24}. $$
A: Use Taylor expansion of $(1+x)^{1/x}$.  
$$(1+x)^{1/x}=e(1- \frac{x}{2} + \frac{11x^2}{24} + ..........)$$
The other terms wont be required in this limit.
The numerator simplifies and the $x$ terms get cancelled out.
A: 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.
A: $(1+x)^{\tfrac 1x}=\mathrm e^{\tfrac{\ln(1+x)}x}$, 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
$$\lim_{x\to o} \frac{(1+x)^{\tfrac1x}-e+\dfrac{ex}{2}}{ex^2}=\lim_{x\to o}\frac{11}{24}+o(1)= \frac{11}{24}.$$
