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?

It's a well-known fact that $$\log(1+x)=x-\dfrac{x^2}2+\dfrac{x^3}3+\underset{x\to 0}{o}(x^3)$$ 
Hence we get $$\dfrac 1x\log(1+x)=1+u(x)$$ where $$u(x)=-\dfrac x2+\dfrac{x^2}3+\underset{x\to 0}{o}(x^2)$$ 
Notice that $$\lim_{x\to 0} u(x)=0$$
We can write $$(1+x)^{1/x}=ee^{u(x)}$$
But we know that $$e^u=1+u+\dfrac{u^2}2+\underset{u\to 0}o(u^2)$$
Now using the identity $$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac$$
we quickly find, since we only need the terms whose degrees are less than three (the other terms are all some $\underset{x\to 0}o(x^2)$), that 
$$(u(x))^2=\left(-\dfrac x2+\dfrac{x^2}3+\underset{x\to 0}{o}(x^2)\right)^2=\dfrac{x^2}4+\underset{x\to 0}o(x^2)$$
Hence, since $\displaystyle\lim_{x\to 0}u(x)=0$, we see than an $\underset{u\to 0}o(u^2)$ function is an $o(x^2)$ when $x$ tends to $0$.
This leads to
$$\begin{align*}(1+x)^{1/x} & = e\left(1+\left(-\dfrac x2+\dfrac{x^2}3+\underset{x\to 0}{o}(x^2)\right)+\dfrac 12\left(\dfrac{x^2}4+\underset{x\to 0}o(x^2)\right)+\underset{x\to 0}{o}(x^2)\right)\\ & = e-\dfrac{ex}2+\dfrac{11e}{24}x^2+\underset{x\to 0}o(x^2)\end{align*}$$
And so :
$$\dfrac{(1+x)^{1/x}-e+\dfrac{ex}2}{x^2}=\dfrac{11e}{24}+\underset{x\to 0}o(1)$$
which means exactly that $$\lim_{x\to 0}\dfrac{(1+x)^{1/x}-e+\dfrac{ex}2}{x^2}=\dfrac{11e}{24}$$
