Let $e$ defined as $\lim\limits_{n\to\infty}(1+\frac{1}{n})^n$.
Let now $f,g:\mathbb{R}\to\mathbb{R}$ defined as: $$f(x)=\sup \{e^r\mid r\in \mathbb{Q}\text{ and } r<x\}$$ And $$g(x)=\lim\limits_{n\to\infty}(1+\frac{x}{n})^n$$ Then $f(x)=g(x)$ for all $x$.
I want to prove this, but I don't know how to start. Any ideas?
On
One way to do this is to define $$L(x) = \int_1^x \frac 1t \, dt, \quad x > 0$$ and show that $f$ is differentiable and monotone on $(0,\infty)$ with range $(-\infty,\infty)$. Define $E(x) = L^{-1}(x)$. It is an exercises in calculus to deduce the usual rules of logarithms and exponents.
Show that $g(x) = E(x)$ by showing that $\lim_{n \to\infty} L((1+x/n)^n) = x$ and invoking continuity of $E$ and $L$.
Show that if $r$ is rational, then $E(r) = e^r$. Deduce that $f(x) = E(x)$ invoking the fact that $E$ is continuous.
If you know that the function $\mathbb{R} \to \mathbb{R}, x\mapsto x^{\frac{p}{q}}$ is continuous for $p,q \in \mathbb{Z}$, $q\ne 0$ then you can rewrite $f$ as $$ f\left( \frac{p}{q} \right)= \left( \lim_{n\to \infty} \left(1+\frac{1}{n}\right)^n \right)^{\frac{p}{q}} = \lim_{n\to \infty} \left(1+\frac{1}{n}\right)^{\frac{np}{q}} .$$
Now you can also look at the subsequence were $n \in q\mathbb{N}$ and replace $n$ by $n'$ using $n'q= n$.
In the formula $$ g\left( \frac{p}{q} \right)= \lim_{n\to \infty} \left(1+\frac{p}{qn}\right)^{n} .$$ you can do the same thing with $n'p=n$.
Now you see that $f$ and $g$ agree on $\mathbb{Q}$. If you know that $g$ is continuous you are done. You can see this by the fact that $g$ is monotonic, the equation $g(r+s)=f(r+s)=f(r)f(s)= g(r)g(s)$ for $r,s \in \mathbb{Q}$ and the fact that $\lim_{r\to 0} g(r) =1$ for $r\in \mathbb{q}$.