Show that $e^{x+y}=e^xe^y$ using $e^x=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n$. I was looking for a proof of $e^{x+y}=e^xe^y$ using the fact that $$e^x=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n.$$
So I have that $$\left(1+\frac{x+y}{n}\right)^n=\sum_{k=0}^n\binom{n}{k}\frac{(x+y)^k}{n^k}=\sum_{k=0}^n\frac{1}{n^k}\sum_{i=0}^k\binom{k}{i}x^iy^{k-i}=\sum_{k=0}^n\binom{n}{k}\sum_{i=0}^k\binom{k}{i}\left(\frac{x}{n}\right)^i\left(\frac{y}{n}\right)^{k-i}$$
But I can't get $$\left(1+\frac{x+y}{n}\right)^n=\sum_{k=0}^n\binom{n}{k}\left(\frac{x}{n}\right)^k\sum_{i=0}^n\binom{n}{i}\left(\frac{y}{n}\right)^i.$$
Any idea ?
 A: \begin{align*}
e^x e^y &= \lim_{n\rightarrow \infty}(1+\frac{x}{n})^n\lim_{n\rightarrow \infty}(1+\frac{y}{n})^n\\
&=\lim_{n\rightarrow \infty}(1+\frac{x}{n})^n(1+\frac{y}{n})^n\\
&=\lim_{n\rightarrow \infty}(1+\frac{x+y}{n}+\frac{xy}{n^2})^n\\
&=\lim_{n\rightarrow \infty}(1+\frac{x+y}{n})^n \frac{(1+\frac{x+y}{n}+\frac{xy}{n^2})^n}{(1+\frac{x+y}{n})^n}\\
&=\lim_{n\rightarrow \infty}(1+\frac{x+y}{n})^n \left(1+\frac{xy}{n^2(1+\frac{x+y}{n})}\right)^n\\
&=\lim_{n\rightarrow \infty}(1+\frac{x+y}{n})^n\left(1+\frac{xy}{n^2(1+\frac{x+y}{n})}\right)^{\frac{n^2(1+\frac{x+y}{n})}{xy}\frac{xy}{n^2(1+\frac{x+y}{n})}n}\\
&= \lim_{n\rightarrow \infty}(1+\frac{x+y}{n})^n\\
&=e^{x+y}.
\end{align*}
A: My favorite approach to these problems is to first prove:

Lemma: If $g$ is a real-valued function on the natural numbers such that $$\lim_{n\to\infty}n(g(n)-1)=c$$ then:
$$\lim_{n\to\infty} g(n)^{n} = e^c$$

Proof:
Case c=0:
Restrict to $n$ so that $n|g(n)-1|<1$. Then:
$$\begin{align}
g(n)^n &= (1+(g(n)-1))^n\\
& = 1+ \sum_{k=1}^{n}\binom{n}{k}(g(n)-1)^k\end{align}$$
Since $\binom{n}{k}\leq \frac{n^k}{k!}$ and $\frac{1}{k!}\leq \frac{1}{2^{k-1}}$:
$$\begin{align}
|g(n)^n-1| &\leq \sum_{k=1}^{n}\frac{1}{k!}\left(n|g(n)-1|\right)^k\\
&\leq n|g(n)-1|\sum_{k=1}^\infty \frac{1}{2^{k-1}}\\
&=2n|g(n)-1|
\end{align}$$
So $\left|g(n)^n-1\right|\to 0$.
General case:
Let $h(n)=\frac{g(n)}{1+c/n}$. Then $$n(h(n)-1)=\frac{n(g(n)-1)-c}{1+c/n}\to 0.$$ So, by the case $c=0$, we have that $h(n)^n\to 1$. But we know that $(1+c/n)^n\to e^c$, so we are done.

The case $c=0$ used only binomial theorem and basic inequalities. The general case requires knowledge that $\lim_{n\to\infty}\left(1+\frac{c}{n}\right)^n\to e^c.$
The lemma can be rewritten in little-$o$ notation as:

If $g(n)=1+\frac{c}{n}+o\left(\frac1n\right)$ then $g(n)^n\to e^c$.


The problem at hand
Next, show that, for fixed $x,y$ that:
$$\begin{align}G(n)&=\left(1+\frac{x}{n}\right)\left(1+\frac{y}{n}\right)
\end{align}$$
satisfies the property that $n(G(n)-1)\to x+y$ since $$n(G(n)-1)=x+y+\frac{xy}{n}.$$
This shows what you wanted.

Another application: Euler's formula
If $h(x)$ is a function such that $h(x)=1+cx+o(x)$ then $h(x/n)^n\to e^{cx}$. This condition $h(x)=1+cx+o(x)$ is equivalent to $h(0)=1,h'(0)=c$.
One interesting case is $h(x)=\cos x+i\sin x$. We can show purely geometrically that $\cos x+i\sin x= 1+ix+o(x)$, and by induction prove that $h(x)^n=h(nx)$, so we have that $$\cos x+i\sin x =h(x)= \lim_{n\to\infty} h(x/n)^n=\lim_{n\to\infty}\left(1+\frac{ix}n\right)^n$$
The geometric proof is roughly, for all $x\in(0,\pi/2)$:

*

*$\sin x<x$. This is because $\frac{1}{2}\sin x$ is the area of a triangle contained in the circle wedge of angle $x$ which has area $x/2$.


*$0\leq 1-\cos x<\frac{1}{2}x^2$. This is true because $1-\cos x = 2\sin^2(x/2)<\frac{x^2}{2}$ by step (1).


*$x<\tan x$. This is because the wedge of angle $x$ is contained in the triangle $(0,0)$, $(1,0)$, $(1,\tan x)$ of area $\frac{1}{2}\tan x$.


*From (3) and (1), we get $x\cos x< \sin x<x$. But $\cos x =1+o(x)$ so $x\cos (x)=x+o(x^2)$ and thus $\sin(x)=x+o(x^2)$.
This shows that $\cos x +i\sin x = 1+ix + o(x)$.
