Evaluation of $\lim_{x\to 0} \frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}$ Evaluate the following limit:
$$L=\lim_{x\to 0} \frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}$$
Using $\ln(1+x)=x-x^2/2+x^3/3-\cdots$
I got $(1+x)^{1/x}=e^{1-x/2+x^2/3-\cdots}$
Could some tell me how to proceed further?

Indeed, the expansion of $(1+x)^{1/x}$ about $x=0$ is $e - \frac{ex}{2} + \frac{11e}{24} x^2 + O(x^3)$, so the limit is then $\dfrac{11e}{24}$.
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To get this expansion, start with
$$(1+x)^{1/x} = \exp \left({1 - x/2 + x^2 / 3 - \dots}\right)$$
which you already found and put this into the power series of $\exp$ to get
$$\sum_{n=0}^{\infty} \frac{(1 - x/2 + x^2/3 - \dots)^n}{n!}$$
The coefficient of $x^0$ is 
$$\sum_{n=0}^{\infty} \frac{1}{n!} = e$$
The coefficient of $x^1$ in $\dfrac{(1-x/2+x^2/3 - \dots)^n}{n!}$ is $-\frac{n/2}{n!}$, hence the coefficient in total is
$$\sum_{n=1}^{\infty}  -\frac{1}{2 (n-1)!} = -\frac{e}{2}$$
I'll leave the coefficient of $x^2$ to you. (It's the hardest, but not too bad).
