real analysis - Is there a way to show that $e^x=lim_{nto infty }left(1+frac{x}{n}right)^n$?

I know that $$e:=\lim_{n\to \infty }\left(1+\frac{1}{n}\right)^n,$$
by definition. Knowing that, I proved successively that $$e^{k}=\lim_{n\to \infty }\left(1+\frac{k}{n}\right)^n,$$
when $k\in \mathbb N$, $k\in \mathbb Z$ and $k\in\mathbb Q$. Now, I was wondering : how can I extend this result over $\mathbb R$ ? I tried to prove that $f_n(x):=(1+\frac{x}{n})^n$ converge uniformly on $\mathbb R$ but unfortunately it failed (I'm not sure that it's even true). Any idea ?

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My idea was to define the function $x\longmapsto e^x$ as $$e^x=\begin{cases}e^x& x\in \mathbb Q\\ \lim_{n\to \infty }e^{k_n}&\text{if }k_n\to x \text{ and }(k_n)\subset \mathbb Q\end{cases}.$$
But to conclude that $$e^x=\lim_{n\to \infty }\left(1+\frac{x}{n}\right)^n,$$
I need to prove that $f_n(x)=\left(1+\frac{x}{n}\right)^n$ converge uniformly on a neighborhood of $x$, but I can't do it. I set $$g_n(x)=f_n(x)-e^x,$$
but I can't find the maximum on a compact that contain $x$, and thus can't conclude.

Answer

We can use that exists $p_n, q_n \in \mathbb{Q}$ such that $p_n,q_n \to x$ and $p_n\le x\le q_n$, therefore

$$\left(1+\frac{p_n}{n}\right)^n\le \left(1+\frac{x}{n}\right)^n\le \left(1+\frac{q_n}{n}\right)^n$$

and

$$\left(1+\frac{p_n}{n}\right)^n=\left[\left(1+\frac{p_n}{n}\right)^\frac{n}{p_n}\right]^{p_n}\to e^x$$

$$\left(1+\frac{q_n}{n}\right)^n=\left[\left(1+\frac{q_n}{n}\right)^\frac{n}{q_n}\right]^{q_n}\to e^x$$

indeed for $\frac{n}{p_n}\in (m,m+1)$ with $m\in \mathbb{N}$ we have

$$\left(1+\frac1{m+1}\right)^m\le \left(1+\frac{p_n}{n}\right)^\frac{n}{p_n}\le \left(1+\frac1m\right)^{m+1}$$

and therefore $\left(1+\frac{p_n}{n}\right)^\frac{n}{p_n}\to e$.