For $e:=\lim\limits_{n\rightarrow\infty}(1+\frac{1}{n})^n$ prove $e^x = \lim\limits_{n\rightarrow\infty} (1+\frac{x}{n})^n$ I couldn't find this exact question, apologize if it's a duplicate.
I would like to show based only on the above definition of e, that this equality for $e^x$ holds, without going through showing it's a bijective function, has an inverse function ln, finding it's derivative and finding it's taylor series.
I've tried to play with it, but except for x=-1 (with a proof that doesn't seem to be possible to generalize), I had no progress.
Is their a straightforward proof for that? Or does one has to go through the whole process mentioned above?
 A: Extending the Limit Over the Integers to the Limit Over the Reals
For $\alpha\ge1$,
$$
\underbrace{\left(1+\frac1{\lfloor\alpha\rfloor+1}\right)^{\large\lfloor\alpha\rfloor}}_{\color{#C00000}{\left(1+\frac{\large1}{\lfloor\alpha\rfloor+1}\right)^{-1}}\color{#00A000}{\left(1+\frac{\large1}{\lfloor\alpha\rfloor+1}\right)^{\large\lfloor\alpha\rfloor+1}}}
\le\left(1+\frac1\alpha\vphantom{\frac1{\lfloor\alpha\rfloor}}\right)^{\large\alpha}
\le\underbrace{\left(1+\frac1{\lfloor\alpha\rfloor}\right)^{\large\lfloor\alpha\rfloor+1}}_{\color{#C00000}{\left(1+\frac{\large1}{\lfloor\alpha\rfloor}\right)}\color{#00A000}{\left(1+\frac{\large1}{\lfloor\alpha\rfloor}\right)^{\lfloor\alpha\rfloor}}}\tag{1}
$$
where the red terms tend to $1$ and the green terms tend to $e$ by the standard limit
$$
\lim_{n\to\infty}\left(1+\frac1n\right)^n=e\tag{2}
$$
where $n\in\mathbb{Z}$. Therefore, $(1)$ and the Squeeze Theorem says that
$$
\lim_{\alpha\to\infty}\left(1+\frac1\alpha\right)^{\large\alpha}=e\tag{3}
$$
where $\alpha\in\mathbb{R}$. Furthermore, $(3)$ implies
$$
\begin{align}
\lim_{\alpha\to\infty}\left(1-\frac1\alpha\right)^{\large-\alpha}
&=\lim_{\alpha\to\infty}\left(1+\frac1{\alpha-1}\right)^{\large\alpha}\\
&=\lim_{\alpha\to\infty}\left(1+\frac1{\alpha-1}\right)^{\large\alpha-1}\left(1+\frac1{\alpha-1}\right)\\[9pt]
&=e\cdot1\tag{4}
\end{align}
$$
Thus, $(3)$ and $(4)$ give
$$
\lim_{|\alpha|\to\infty}\left(1+\frac1\alpha\right)^{\large\alpha}=e\tag{5}
$$

Applying $\boldsymbol{(5)}$ to the Question
$$
\begin{align}
\lim_{n\to\infty}\left(1+\frac xn\right)^n
&=\lim_{n\to\infty}\left(1+\frac xn\right)^{\large\frac nx\cdot x}\\[9pt]
&=e^x\tag{6}
\end{align}
$$
for all $x$.
A: Here, by substitution...
$$\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n = \lim_{n\to\infty}\left(1+\frac{1}{\frac{n}{x}}\right)^n = \lim_{m\to\infty}\left(1+\frac{1}{m}\right)^{xm} = \lim_{m\to\infty}\left(\left(1+\frac{1}{m}\right)^m\right)^x = e^x$$
This will only work for positive x, but if you've already got it for -1, you should be able to take it from there.
