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.

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...

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 $-e/2$.