Definition of $\pi$, $\lim\limits_{n \to \infty}{n \sin\left(\frac{180^o}{n}\right)}$ I'm learning mathematical analysis recently. My book gave the definition of $\pi$ as the limit of sequence $\left\{n \sin \frac{180^o}{n}\right\}$.
The way it prove this sequence is convergent is quite strange to me. It first showed the sequence is smaller than 4, then monotonically increasing.
The latter part is confusing. It first let $t = \frac{180^o}{n(n+1)} $, and proved $\tan nt \ge n\tan t$ for $nt \le 45^o$, so
$$
\sin(n+1)t = \sin nt \cos t + \cos nt \sin t
= \sin nt \cos t\left(1 + \frac{\tan t}{\tan nt}\right)
\le \frac{n+1}{n} \sin nt
$$
then
$$
n \sin \frac{180^o}{n} \le (n+1) \sin \frac{180^o}{n+1}
$$
This is perfectly correct, but how can I come up with a $t$ like this? If I'm to prove this, is there a way to figure out what the $t$ should be like? Or all I can do is just memorize it? Alternately, do you guys have a more intuitive proof?
 A: A more intuitive proof could look as follows:


*

*Set $n=1/x$, then you have $\lim \limits_{x\to 0} \frac{\sin(ax)}{x} $.

*Now do a series expansion to get $\lim \limits_{x\to 0} a-\frac{a^3x^3}{6} +\dots =a$

*Plugin in $a=\pi$, if you like ...

A: Here's a more intuitive proof:
Take an $n$-sided polygon. Radius (center to vertex) $R$. Calculate the perimeter of the polygon. It will be:
$$P= 2\cdot R\cdot n\cdot \sin{ \theta}$$
where $\theta$ is $\frac{360}{2n}$ (Get this by dropping a perpendicular from vertex to a side, calculate the length of the side, remembering that we've bisected $\theta$ and the side. Multiply by $n$.)
Now watch this polygon become a circle:
$$P=\lim\limits_{n \rightarrow \infty} 2\cdot R\cdot n\cdot \sin\left(\frac{180}{n}\right) = 2\cdot \left(\lim\limits_{n \rightarrow \infty} n\cdot \sin\left(\frac{180}{n}\right)\right)\cdot R$$
But it has to be $2\cdot \pi\cdot R$. So $\lim\limits_{n \rightarrow \infty} n\cdot \sin\left(\frac{180}{n}\right)$ must be $\pi$.
A: Here's a pretty simple proof I know for your problem:
$$\lim_{x\to\infty}{\left[x\cdot\sin{\frac{a}{x}}\right]}$$
Let $$\frac{a}{x}=u$$
$$\Leftrightarrow\lim_{x\to\infty}{\left[x\cdot\sin{\frac{a}{x}}\right]}=\lim_{\frac{a}{u}\to\infty}{\left[\frac{a}{u}\cdot\sin{u}\right]}$$
With $$\frac{a}{u}\to\infty\Leftrightarrow u\rightarrow 0$$
$$\Leftrightarrow\lim_{x\to\infty}{\left[x\cdot\sin{\frac{a}{x}}\right]}=a\cdot\lim_{u\to 0}{\left[\frac{\sin{u}}{u}\right]}$$
There is a theorem that says: $$\lim_{u\to 0}{\left[\frac{\sin{u}}{u}\right]}=1$$
$$\Leftrightarrow\lim_{x\to\infty}{\left[x\cdot\sin{\frac{a}{x}}\right]}=a$$
q.e.d.
(Please note that $180°=\pi$. That's why $\Leftrightarrow\lim_{x\to\infty}{\left[x\cdot\sin{\frac{180°}{x}}\right]}=\pi$)
So notice that not any of the proofs here, not even yours, can be used to define pi, as a definition of something cannot contain that thing itself.
It's like saying: "documentation is a word that means documentation"
A: I actually did this by accident the other day: Using the area formula for a circle $\pi r^2$ and a simplified version of the area of a regular polygon $r^2  n  sin(180/n)  cos(180/n)$, I cancelled out the squares radio and thought about it as $n$ approaches infinity. I found I could ignore $cos$, as $\underset{n \to \infty}{lim}cos (180/n) = 1$, which is safe to ignore during multiplication, and I ended up with this:
$\pi = n  sin(180/n)$
As $n$ approaches infinity, $n sin(180/n)$ approaches $\infty  \times 0$, which is something I as a high schooler was very happy to see. 
A: A simpler way.
Imagine a circle as an equilateral polygon with $n$ sides. And to make things easier, imagine the distance from polygon edges to the centre as the unit $1$ (kinda like a unit circle). Let's call it a "unit equilateral polygon".
Now, let's cut the polygon into $n$ small triangles like how you cut a cake.
Now, each of these triangles has two sides of unit $1$ and an angle of $\frac{360^\circ}{n}$ between them.
Let's cut this triangle itself into two right angle halves. The angle is now $\frac{180^\circ}{n}$. Notice the opposite of this angle is simply $\sin(\frac{180^\circ}{n})$
Multiply this by $2$ to get the length of one side of the polygon $2 \space \sin(\frac{180^\circ}{n})$.
Remember, there are $n$ of these sides. So
Circumference of n-equilateral polygon $= 2 \space n \space \sin(\frac{180^\circ}{n})$
Now, given a circle is nothing but an equilateral polygon with an infinite number of sides.
$\text{Unit Circle Circumference} = 2 \Pi = \lim_{n \to \infty} 2 \space n \space \sin(\frac{180^\circ}{n})$
we divide by $2$ to get
$\Pi = \lim_{n \to \infty} \space n \space \sin (\frac{180^\circ}{n})$
Job done.
