Need hint for $\lim_{x\to 0} \frac{(x+1)^\frac{1}{x}-e}{x}$ [duplicate]

Please give me some hints or solution for this limit. I had it on my exam and I'm curious how to solve it. $$ \lim_{x\to 0} \frac{(x+1)^\frac{1}{x}-e}{x} $$

Easiest way is by series expansion: $$\frac{1}{x}\log(1+x) = \frac{1}{x} \left( x - \frac{x^2}{2} + O(x^3) \right) = \left( 1 - \frac{x}{2} + O(x^2) \right) $$ $$ (x+1)^{1/x} = e^{\frac{1}{x}\log(1+x) } = e^{\left( 1 - \frac{x}{2} + O(x^2) \right)} = \sum_{n=0}^\infty \frac{\left( 1 - \frac{x}{2} + O(x^2) \right)^n}{n!} $$ $$ (x+1)^{1/x}-e = \sum_{n=0}^\infty \frac{\left( 1 - \frac{x}{2} + O(x^2) \right)^n}{n!} - \sum_{n=0}^\infty \frac{1}{n!} = \sum_{n=0}^\infty \frac{\left( 1 - \frac{x}{2} + O(x^2) \right)^n-1}{n!} $$ $$ \frac{(x+1)^{1/x}-e}{x} = \sum_{n=0}^\infty \frac{\left( 1 - n\frac{x}{2} + O(x^2) \right)-1}{xn!} = \sum_{n=0}^\infty \frac{ - n\frac{x}{2} + O(x^2) }{xn!} $$ $$ \frac{(x+1)^{1/x}-e}{x} = -\frac{1}{2} \sum_{n=0}^\infty \frac{n + O(x)}{n!} = -\frac{1}{2} \sum_{n=0}^\infty \frac{n }{n!} +O(x) $$ $$ \frac{(x+1)^{1/x}-e}{x} = -\frac{1}{2} \sum_{n=1}^\infty \frac{n}{n!} +O(x) = -\frac{1}{2} \sum_{m=0}^\infty \frac{1}{(m+1)!} +O(x) = -\frac{e}{2} + O(x) $$ from which it follows that $$ \lim_{x\rightarrow \infty}\frac{(x+1)^{1/x}-e}{x} = -\frac{e}{2} $$

Use l'Hospital, since $\;(x+1)^{1/x}\xrightarrow[x\to 0]{}e\;$ , writing

$$(x+1)^{1/x}=e^{\frac1x\log(x+1)}\implies\;\text{the limit thus is} \;\;\lim_{x\to 0}\frac{e^{\frac1x\log(x+1)}-e}x\;:$$

$$\lim_{x\to 0}\frac{e^{\frac{\log(x+1)}x}-e}x\stackrel{\text{l'H}}=\lim_{x\to 0}\frac{x-(x+1)\log(x+1)}{x^2(x+1)}e^{\frac{\log(x+1)}x}\stackrel{\text{l'H}}=-\frac e2$$

Hint: Consider the function $f(x) = (x + 1)^{1/x}$ for $x \neq 0$ and $f(0) = e$. Then $f$ is right-continuous at everywhere and the limit you are asked to find is the right-derivative of $f$ at $x = 0$, if it exists. So find $f'$ and evaluate it at $x = 0$.