Definition extension of $e$ to the reals I have read proofs that the sequence $a_n=\left( 1+\frac{1}{n}\right)^n,n\in\mathbb{N^*}$ converges and its limit is defined to be $e$.
How is this definition of $e$ extended to the real numbers?
Phrased in another way: How can we prove the following:

Letting $f(x) = \left( 1+\frac{1}{x} \right) ^x$, prove that
  $\lim_\limits{x\to +\infty}{f(x)}=e$.

This has to be proven only with the definition that $e=\lim_\limits{n\to +\infty}{\left[ \left( 1+\frac{1}{n}\right)^n\right] }, n\in\mathbb{N^*}$, for the extension to be correct.

Let $n=\lfloor x\rfloor$, then:
$$
\left(1+{1\over n+1}\right)^n \le 
\left(1+{1\over x}\right)^x \le
\left(1+{1\over n}\right)^{n+1},
$$
that is
$$
\tag{1}
{\left(1+{1\over n+1}\right)^{n+1}\over \left(1+{1\over n+1}\right)} \le 
\left(1+{1\over x}\right)^x \le
\left(1+{1\over n}\right)^{n}\left(1+{1\over n}\right).
$$
The left and right hand side of this inequality can be viewed as two sequences, $a_n$ and $b_n$, both converging to $e$ (by definition of $e$ and standard theorems on the limit of a product of two sequences). So for any $\epsilon>0$ there exists $M$ such that $e-\epsilon<a_n$ and $b_n<e+\epsilon$ when $n>M$.
In virtue of $(1)$ we have then $e-\epsilon< \left(1+{1\over x}\right)^x <e+\epsilon$ for all $x>M+1$, that is: $\lim\limits_{x\to\infty}\left(1+{1\over x}\right)^x=e$.
