Limit as $x\to 0$ of $\frac{(1+x)^{1/x}-e}{x}$ Find the following limit:
$$\lim_{x\to 0}\frac{(1+x)^{1/x}-e}{x} \tag{1}$$
The following is my approach, although is full of incorrect assumptions (statements etc). 
$$f(x)= \lim_{h\to 0}(1+x+h)^{1/(x+h)}\\$$
From here we can say $f(0) = e$.
$$\ln(f(x)) = \lim_{h\to 0}\frac{1}{x+h}\ln(1+x+h)$$
Near $x=0$, we can use the series of logarithm:
$$\ln(f(x)) = \lim_{h\to 0}\frac{1}{x+h}\left(x+h-\frac{(x+h)^2}{2}...\right)$$
Differentiating we get:
$$\frac{f'(x)}{f(x)} = \lim_{h\to 0}-\frac{1}{2} +\frac{x+h}{3}  ... \tag{2}$$
Now we note that $(1)$ is actually $f'(0)$ (?) and so from $(2)$ we get:
$$f'(0) =\frac{-1}{2} f(0) = \frac{-e}{2}$$
While the answer is seemingly correct, the method is absolutely not. It looks like this is fluke than anything else.
I also tried computing taylor series of $(1+x)^{1/x}$ near $x=0$ but couldn't do it.
 A: You are making things complicated by bringing in the $h$. One has
$$\ln[(1+x)^{1/x}]=1-\frac x2+O(x^2)$$
so taking exponentials gives
$$(1+x)^{1/x}=e\exp(-x/2+O(x^2))=e(1-x/2+O(x^2)).$$
Then
$$\frac{(1+x)^{1/x}-e}{x}=-\frac e2+O(x)$$
etc.
This is really the same sort of manipulation as you were doing...
A: Why don't you just take $e$ as a factor and transform the given expression into $$e\cdot\dfrac{\exp\left(\dfrac{\log(1+x)}{x}-1\right) - 1} {\dfrac{\log(1+x)}{x}-1}\cdot\frac{\log(1+x)-x}{x^2} $$ The middle factor tends to $1$ because argument of $\exp$ tends to $0$. The last factor tends to $-1/2$ via Taylor series or L'Hospital's Rule so that the final answer is $-e/2$.
A: Hint
Consider $$y=(1+x)^{1/x}\implies \log(y)=\frac 1x \log(1+x)$$ Now, using Taylor
$$\log(y)=\frac 1x\left(x-\frac{x^2}{2}+\frac{x^3}{3}+O\left(x^4\right) \right)=1-\frac{x}{2}+\frac{x^2}{3}+O\left(x^3\right)$$ Now $$y=e^{\log(y)}=e-\frac{e x}{2}+\frac{11 e x^2}{24}+O\left(x^3\right)$$ Just continue.
Edit
For the limit itself, the development to $O\left(x^2\right)$ was enough. Doing it to $O\left(x^3\right)$ allows to find the limit and also how it is approached.
