<s>[INST] Is there a way to show that $e^x=\lim_{n\to \infty }\left(1+\frac{x}{n}\right)^n$? 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 ?

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.
 [/INST] Left side:
The exponential function may be written as a Taylor series:
$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$
Right side:
$(1+\frac{x}{n})^n$ is a binomial expansion like:
$(1+y)^n=\binom{n}{0}y^0+\binom{n}{1}y^1+\binom{n}{2}y^2+...+\binom{n}{n-1}y^{n-1}+\binom{n}{n}y^n$
Where $\binom{n}{k}$ is the Binomial coefficient given by the formula :
$\binom{n}{k}=\frac{n!}{k!(n-k)!}$
Some basic properties of $\binom{n}{k}$:
a)$\binom{n}{0}=1$ because $\frac{n!}{0!(n-0)!}=\frac{n!}{1*n!}$
b)$\binom{n}{1}=n$ because $\frac{n!}{1!(n-1)!}=\frac{(n-1)!*n}{(n-1)!}$
c)$\binom{n}{n-1}=n$ because $\frac{n!}{(n-1)!(n-(n-1))!}=\frac{(n-1)!*n}{(n-1)!*1!}$
d)$\binom{n}{n}=1$ because $\frac{n!}{n!(n-n)!}=\frac{1}{1!}$
e) The formula does exhibit a symmetry that is less evident from the multiplicative formula:
$\binom{n}{k}=\binom{n}{n-k}$
Returning:
$(1+\frac{x}{n})^n=1+n*\frac{x}{n}+\frac{n!}{2!(n-2)!}\frac{x^2}{n^2}+\frac{n!}{3!(n-3)!}\frac{x^3}{n^3}+...+\frac{n!}{3!(n-3)!}\frac{x^{n-3}}{n^{n-3}}+\frac{n!}{2!(n-2)!}\frac{x^{n-2}}{n^{n-2}}+n*\frac{x^{n-1}}{n^{n-1}}+\frac{x^n}{n^n}$
$(1+\frac{x}{n})^n=1+x+\frac{(n-1)n}{n^2}\frac{x^2}{2!}+\frac{(n-2)(n-1)n}{n^3}\frac{x^3}{3!}+...+\frac{(n-2)(n-1)n}{3!}\frac{x^{n-3}}{n^{n-3}}+\frac{(n-1)n}{2!}\frac{x^{n-2}}{n^{n-2}}+\frac{x^{n-1}}{n^{n-2}}+\frac{x^n}{n^n}$
$(1+\frac{x}{n})^n=1+x+\frac{n-1}{n}\frac{x^2}{2!}+\frac{(n-2)(n-1)}{n^2}\frac{x^3}{3!}+...+\frac{(n-2)(n-1)}{n^{n-4}}\frac{x^{n-3}}{3!}+\frac{n-1}{n^{n-3}}\frac{x^{n-2}}{2!}+\frac{x^{n-1}}{n^{n-2}}+\frac{x^n}{n^n}$
Let's analyze what happens for $n\rightarrow\infty$-here we have three types of limits:
-First type:
$\displaystyle\lim_{n \to \infty}\frac{n-1}{n}=\displaystyle\lim_{n \to \infty}[1+\frac{1}{n}]=1+0=1$
$\displaystyle\lim_{n \to \infty}\frac{(n-2)(n-1)}{n^2}=\displaystyle\lim_{n \to \infty}\frac{n^2-3n+2}{n^2}=\displaystyle\lim_{n \to \infty}[1-\frac{3}{n}+\frac{2}{n^2}]=1-0+0=1$
Hense $\displaystyle\lim_{n \to \infty}\Bigg(\frac{\displaystyle\prod_{i=1}^{k} (n-i)}{n^k}\Bigg)=1$
-Second type is
$\displaystyle\lim_{n \to \infty} \frac{x^{n-\alpha}}{n^{n-\beta}}$-Because ${n^{n-\beta}}$ grows much fasten than $x^{n-\alpha}$ hense: $\displaystyle\lim_{n \to \infty} \frac{x^{n-\alpha}}{n^{n-\beta}}=0$
-Third type:
$\displaystyle\lim_{n \to \infty}\Bigg(\frac{\displaystyle\prod_{i=1}^{k} (n-i)}{n^{n-k-1}}\frac{x^{n-k}}{k!}\Bigg)$
We have to show on the biggest power (similar to the first type) as the most relevant:
$\frac{\displaystyle\prod_{i=1}^{k} (n-i)}{n^{n-k-1}}\frac{x^{n-k}}{k!}\sim\frac{n^{k-1} }{n^{n-k-1}}\frac{x^{n-k}}{k!}=n^{k-1-(n-k-1)}\frac{x^{n-k}}{k!}=n^{2k-n}*\frac{x^{n-k}}{k!}=\frac{1}{k!}*\frac{x^{n-k}}{n^{n-2k}}$
Again: ${n^{n-\beta}}$ grows much faster than $x^{n-\alpha}$
Hense: $\displaystyle\lim_{n \to \infty}\Bigg(\frac{\displaystyle\prod_{i=1}^{k} (n-i)}{n^{n-k-1}}\frac{x^{n-k}}{k!}\Bigg)=0$
Our right side equals:
$\displaystyle\lim_{n \to \infty}(1+\frac{x}{n})^n=1+x+1*\frac{x^2}{2!}+1*\frac{x^3}{3!}+...+0+0+0+0$
$\displaystyle\lim_{n \to \infty}(1+\frac{x}{n})^n=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$
We got the same elements like in the Taylor series of $e^x$. Q.E.D.
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