The task is to show why e^2x = (1+2x/n)^n for very large n. I'm not sure how to tackle this problem. I thought about creating the generating functions and doing some algebraic manipulation but couldn't finish it.
The sequence $\displaystyle \displaystyle a_n=\left(1+\frac{2x}{n}\right)^n$ converges to $\displaystyle e^{2x}$ when $\displaystyle n\to\infty$
$\displaystyle \displaystyle \lim_{n\to\infty}\left(1+\frac{2x}{n}\right)^n=\li m_{n\to\infty}\left[\left(1+\frac{2x}{n}\right)^{\frac{n}{2x}}\right]^{\frac{2x}{n}\cdot n}=e^{2x}$
$\displaystyle \displaystyle\lim_{n\to\infty}\left(\cos\frac{x}{\ sqrt{n}}\right)^n=\lim_{n\to\infty}\left(1+\cos\fr ac{x}{\sqrt{n}}-1\right)^n=\lim_{n\to\infty}\left(1-2\sin ^2\frac{x}{2\sqrt{n}}\right)^n=$
$\displaystyle \displaystyle =e^{\displaystyle -\lim_{n\to\infty}2\sin ^2\frac{x}{2\sqrt{n}}\cdot n}=e^{\displaystyle -\lim_{n\to\infty}\left(\frac{\sin \frac{x}{2\sqrt{n}}}{\frac{x}{2\sqrt{n}}}\right)^2 \cdot\frac{x^2}{2}}=e^{-\displaystyle \frac{x^2}{2}}$
Another method is to define:
$\displaystyle
f_n(x)=(1+2x/n)^n
$
then
$\displaystyle
f'_n(x)=2 f_n(x) (1+O(x/n))
$
which is of seperable type so:
$\displaystyle
\int \frac{1}{f_n} df_n = \int (2+O(x/n)) dx
$
or:
$\displaystyle
f_n(x)= Ae^{2x+O(x^2/n)}
$
and as $\displaystyle f_n(0)=1,\ A=1$ (though a bit of care may be needed to show this)
Hence for fixed $\displaystyle x$ we have:
$\displaystyle
\lim_{n \to \infty} f_n(x) = e^{2x}
$
RonL
Here's a fairly simple proof I just scratched down.
$\displaystyle \lim_{n\rightarrow{\infty}}\left(1+\frac{2x}{n}\ri ght)^{n}$
Let $\displaystyle a=\frac{2x}{n}, \;\ n=\frac{2x}{a}$
$\displaystyle \lim_{a\rightarrow{0}}\left(1+a\right)^{\frac{2x}{ a}}$
$\displaystyle \lim_{a\rightarrow{0}}[\underbrace{\left(1+a\right)^{\frac{1}{a}}}_{\text {limit=e}}]^{2x}=e^{2x}$