Show that $e^x=\lim_{n\to\infty }(1+\frac{x}{n})^n$ using the fact that $e^q=\lim_{n\to\infty}(1+\frac{q}{n})^n$ for $q\in\mathbb Q$. I have to show that $$\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n=e^x$$ for all $x\in\mathbb R$ using the fact that $$e=\lim_{n\to\infty }\left(1+\frac{1}{n}\right)^n.$$
I already showed that for all $q\in\mathbb Q$, $$\lim_{n\to\infty }\left(1+\frac{q}{n}\right)^n=e^q,$$
but I have problem to show the relation for $x\in\mathbb R$. My idea is to take a sequence of rational $(x_n)$ that converge to $x\in\mathbb R\backslash \mathbb Q$ (which exist), and thus
$$\lim_{n\to\infty }\left(1+\frac{x_m}{n}\right)^n=e^{x_m}\implies \lim_{m\to\infty }\lim_{n\to\infty }\left(1+\frac{x_m}{n}\right)^n=\lim_{m\to\infty }e^{x_m},$$
but now, how can I justify that $$\lim_{m\to\infty }\lim_{n\to\infty }\left(1+\frac{x_m}{n}\right)^n=\lim_{n\to\infty }\lim_{m\to\infty }\left(1+\frac{x_m}{n}\right)^n$$ and $$\lim_{m\to\infty }e^{x_m}=e^{\lim_{m\to\infty }x_m}.$$
(I can't use continuity since I'm not supposed to know that $f:x\mapsto e^x$ is continuous).
 A: 
NOTE
We will present an outline of the proof for the case where $x\ge x_m$ for all $m$ and leave the both the details of the proof and development of the case for $x\le x_m$ to the reader.


First, we form a sequence $x_m$ for which $x_m\le x$ for all $m$, and $x_m\to x$.  
Then, for any given $\epsilon>0$, there exists a number $M$ such that for all $m>M$, $0\le x-x_m <\epsilon$.
Next, we write
$$\begin{align}
0\le \left|\left(1+\frac{x}{n}\right)^n-\left(1+\frac{x_m}{n}\right)^n\right|&=\left(1+\frac{x_m}{n}\right)^n\left(\frac{\left(1+\frac{x}{n}\right)^n}{\left(1+\frac{x_m}{n}\right)^n}-1\right)\\\\
&=\left(1+\frac{x_m}{n}\right)^n\left(\frac{1}{\left(1-\frac{x-x_m}{n+x}\right)^n}-1\right) \tag 1\\\\
&\le\left(1+\frac{x_m}{n}\right)^n\left(\frac{1}{\left(1-\frac{n(x-x_m)}{n+x}\right)}-1\right) \tag 2\\\\
&=\left(1+\frac{x_m}{n}\right)^n\left(\frac{n}{n(1-(x-x_m))+x}\right)(x-x_m) \tag 3
\end{align}$$
where in going from $(1)$ to $(2)$, we used Bernoulli's Inequality.
Now, with $m>M$ held fixed, but $M$ chosen so large that $x-x_m<1$, we take $n\to \infty$ in $(3)$.  Then, we have for all $\epsilon>0$ there is an $M$ so that 
$$\left|\left(1+\frac{x}{n}\right)^n-\left(1+\frac{x_m}{n}\right)^n\right|\le e^{x_m}\frac{(x-x_m)}{1-(x-x_m)}<\left(\frac{e^{x_m}}{1-(x-x_m)}\right)\,\epsilon$$
and we are done!
A: Step 1. If $x<y$, then $\displaystyle\left(1+\frac{x}{n}\right)^n<\left(1+\frac{y}{n}\right)^n$ - Straightforward.
Step 2. Assuming that $\,\lim_{n\to\infty}\left(1+\frac{q}{n}\right)^n=\mathrm{e}^q$, for all $q\in\mathbb Q$, then using Step 1, for every $p<x<q$, $p,q\in\mathbb Q$ and $x\in\mathbb R$, we have that
$$
\mathrm{e}^p\le\liminf_{n\to\infty}\left(1+\frac{x}{n}\right)^n =:a(x)
\le b(x):=\limsup_{n\to\infty}\left(1+\frac{x}{n}\right)^n
\le\mathrm{e}^q.
$$
Step 3. Using the fact that $f(x)=\exp(x)$ is continuous and increasing, let $x\in\mathbb R$, and let $\{p_n\}_{n\in\mathbb N},\{q_n\}_{n\in\mathbb N}\subset\mathbb Q$, $\{p_n\}_{n\in\mathbb N}$ increasing, 
$\{q_n\}_{n\in\mathbb N}$ decreasing, such that $p_n,q_n\to x$, then as 
$\exp(x)$ is continuous and increasing $\lim_{n\to\infty}\mathrm{e}^{p_n}=\lim_{n\to\infty}\mathrm{e}^{q_n}=\mathrm{e}^x$
and
$$
\mathrm{e}^{p_n} < \mathrm{e}^x<\mathrm{e}^{q_n}
\quad\text{and}\quad
\mathrm{e}^{p_n}\le\liminf_{n\to\infty}\left(1+\frac{x}{n}\right)^n =:a(x)
\le b(x):=\limsup_{n\to\infty}\left(1+\frac{x}{n}\right)^n
\le\mathrm{e}^{q_n}.
$$
Hence $a(x)=b(x)=\mathrm{e}^x$, and thus the limit $\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n$ exists and it is equal to $\mathrm{e}^x$.
A: Just an other way
Let $f_n(x)=\left(1+\frac{x}{n}\right)^n$. The $f_n$ are continuous and converge uniformly to $e^x$ on all compact set. Therefore, $x\mapsto e^x$ is continuous on all compact set and thus 
$$e^x\underset{continuity}{=}\lim_{m\to\infty }e^{x_m}=\lim_{m\to\infty }\lim_{n\to\infty }f_n(x_m)\underset{f_n\ converge\ uniformly}{=}\lim_{n\to\infty }\lim_{m\to\infty }f_n(x_m)\underset{continuity}{=}\lim_{n\to\infty }f_n(x)=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n.$$
Do you understand ?
