definition of the constant $e$ To my knowledge there are two possible ways to define $e^x$
$$e^x = \sum_{i=0}^{\infty}\frac{x^i}{i!}$$
$$e^x = \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n$$
So my question is: Why does…
$$\sum_{i=0}^{\infty}\frac{x^i}{i!} = \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n$$
 A: HINT :
Use the General Binomial Theorem on the expression on the right hand side.
A: I'm not giving you the full formal proof - you can look that up in just about any calculus textbook - but the basic idea is to use $$
  (a + b)^n = \sum_{k=0}^n \binom{n}{k} a^k b^{n-k} \text{ where } \binom{n}{k} = \frac{n!}{k!(n-k)!}
$$
to get $$
  \left(1 + \frac{x}{n}\right)^n = \sum_{k=0}^n \frac{n!}{n^kk!(n-k)!}x^k
  = \sum_{k=0}^n \frac{n(n-1)\ldots(n-k+1)}{n^k} \frac{x^k}{k!} \text{.}
$$
Then, if $n >> k$ (meaning $n$ is much larger than $k$), you have $$
  \frac{n(n-1)\ldots(n-k+1)}{n^k} \approx 1 \text{.}
$$
Thus, if you fix $K$, and let $n \to \infty$, the terms involving $x^0,\ldots,x^K$ in $\left(1 + \frac{x}{n}\right)^n$ will converge towards $\frac{x^k}{k!}$.
A: The power series
$$
\sum_{n\ge0}\frac{x^n}{n!}
$$
defines a function on the whole real line, let's call it “exp”. Since power series can be differentiated term by term, we see that $\exp'=\exp$ and also that $\exp0=1$. Consider now the function
$$
f(x)=\exp(a-x)\exp x.
$$
We have
$$
f'(x)=-\exp(a-x)\exp x+\exp(a-x)\exp x=0
$$
so that $f'(x)=0$. Therefore $f$ is constant and its value is $f(0)=\exp a$. So we have
$$
\exp(a-b)\exp b=\exp a
$$
for all real $a$ and $b$. If $a-b=x$ and $b=y$, we can write the relation as
$$
\exp(x+y)=\exp x\exp y
$$
and, in particular $\exp(-x)=(\exp x)^{-1}$. So the function $\exp$ never has the value $0$ and so it's always positive. Therefore it's increasing, hence invertible. Denote by $\log$ its inverse. By the inverse function theorem, we have
$$
\log'1=1
$$
that is
$$
\lim_{h\to0}\frac{\log(1+h)-\log1}{h}=1.
$$
However, $\log1=0$ by definition, so, for any $x$,
$$
x=\lim_{h\to0}x\frac{\log(1+h)}{h}
$$
and, with the substitution $n=x/h$ we get
$$
\lim_{n\to\infty}n\log\left(1+\frac{x}{n}\right)=x.
$$
For natural $n$, it's clear that $n\log t=\log t^n$ (induction on $\log a+\log b=\log(ab)$ which is the same as $\exp(x+y)=\exp x\exp y$), so we have
$$
\lim_{n\to\infty}\log\left(\left(1+\frac{x}{n}\right)^n\right)=x
$$
which amounts to say that
$$
\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=\exp x.
$$
