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?
 A: 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).
A: You can proceed in the following manner
\begin{align}
L &= \lim_{x \to 0}\dfrac{(1 + x)^{1/x} - e + \dfrac{ex}{2}}{x^{2}}\notag\\
&= e\lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + x)}{x} - 1\right) - 1 + \dfrac{x}{2}}{x^{2}}\notag\\
&= e\lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + x)}{x} - 1\right) - \exp\left(\log\left(1 - \dfrac{x}{2}\right)\right)}{x^{2}}\notag\\
&= e\lim_{x \to 0}\exp\left(\log\left(1 - \dfrac{x}{2}\right)\right)\cdot\dfrac{\exp\left(\dfrac{\log(1 + x)}{x} - 1 - \log\left(1 - \dfrac{x}{2}\right)\right) - 1}{x^{2}}\notag\\
&= e\lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + x)}{x} - 1 - \log\left(1 - \dfrac{x}{2}\right)\right) - 1}{\dfrac{\log(1 + x)}{x} - 1 - \log\left(1 - \dfrac{x}{2}\right)}\cdot\dfrac{\dfrac{\log(1 + x)}{x} - 1 - \log\left(1 - \dfrac{x}{2}\right)}{x^{2}}\notag\\
&= e\lim_{x \to 0}\dfrac{\dfrac{\log(1 + x)}{x} - 1 - \log\left(1 - \dfrac{x}{2}\right)}{x^{2}}\notag\\
&= e\lim_{x \to 0}\dfrac{\left(1 - \dfrac{x}{2} + \dfrac{x^{2}}{3} + o(x^{2})\right) - 1 + \left(\dfrac{x}{2} + \dfrac{x^{2}}{8} + o(x^{2})\right)}{x^{2}}\text{ (using Taylor series)}\notag\\
&= e\left(\frac{1}{3} + \frac{1}{8}\right)\notag\\
&= \frac{11e}{24}\notag
\end{align}
The above approach uses the standard limit $$\lim_{t \to 0}\frac{\exp(t) - 1}{t} = 1$$ Taylor series is used only when necessary and this approach avoids any multiplication/division of infinite series.
A: 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 $$\lim_{x\to 0}\dfrac{(1+x)^{1/x}-e+\dfrac{ex}2}{x^2}=\dfrac{11e}{24}$$
