TITLE: Intuitive proofs that $\lim\limits_{n\to\infty}\left(1+\frac xn\right)^n=e^x$
QUESTION [38 upvotes]: At this link someone asked how to prove rigorously that
$$
\lim_{n\to\infty}\left(1+\frac xn\right)^n = e^x.
$$
What good intuitive arguments exist for this statement?  
Later edit: . . . where $e$ is defined as the base of an exponential function equal to its own derivative.
I will post my own answer, but that shouldn't deter anyone else from posting one as well.

REPLY [2 votes]: Starting from the definition $\ln(x) = \int_1^x \frac{dt}{t}$ and $e^{\ln x} = x$ for any $x > 0$.

Let $P_n(x)$ be a sequence of polynomials such that $\frac{P'_n(x)}{P_n(x)} \to 1$ uniformly on any interval $[-A,A]$ and $P_n(0) = 1$. Then $P_n(x) \to e^x$ uniformly on any interval $[-A,A]$.
proof : note that $(\ln P_n(x))' = \frac{P_n'(x)}{P_n(x)}$ so that $\ln P_n(x) = \ln P_n(0) + \int_0^x \frac{P'_n(t)}{P_n(t)} dt \to x$ uniformly, hence $P_n(x) \to e^{x}$ uniformly.


Consider $P_n(x) = (1+\frac{x}{n})^{n}$. $ \ $ Thus $P_n'(x) =  (1+\frac{x}{n})^{n-1}$ so that $\frac{P_n'(x)}{P_n(x)} = \frac{1}{1 +\frac{x}{n}} \to 1$ uniformly on any interval $[-A,A]$. And since $P_n(0) = 1$ : $$\textstyle P_n(x) = \ \ \color{red}{\left(1+\frac{x}{n}\right)^{n} \ \ \to \ \ e^x} \qquad\qquad \text{uniformly on } [-A,A]$$
$$ $$
Consider $P_n(x) = \sum_{k=0}^n \frac {x^k}{k!}$.  $ \ $  Thus  $P_n'(x) = \sum_{k=0}^{n-1} \frac {x^k}{k!} = P_n(x) - \frac{x^n}{n!}$ so that $\frac{P_n'(x)}{P_n(x)} = \frac{P_n(x)}{P_n(x)-\frac{x^n}{n!}} \to 1$ uniformly on any interval $[-A,A]$. And since $P_n(0) = 1$ :
$$\textstyle  P_n(x) = \ \  \color{red}{\sum_{k=0}^n \frac {x^k}{k!} \ \ \to \ \ e^x} \qquad\qquad \text{uniformly on } [-A,A]$$