$\lim_{n \to \infty}n\sin\left(\pi/n\right)$ Assume unit radius. The area of regular polygon with $n$ sides = $n\sin\left(\pi/n\right)$, how does $n\sin\left(\pi/n\right)$ approach $1$ as $n \to \infty$ ?.
$$
\mbox{I have tried}\quad
\pi\,{\sin\left(\pi/n\right) \over \pi/n}\quad \mbox{so it looks like}\quad \pi\,{\sin(x) \over x}
$$
L'Hopital's Rule and I still can not get
$\quad\lim_{x \to \infty}\sin\left(x\right)/x = 1$.
 A: As $\epsilon\to0$, we have $\sin\epsilon=\epsilon+o(\epsilon)$ (it's Taylor's formula around $0$, at order $1$), hence, as $n\to\infty$,
$$n\sin\frac\pi n=n\left(\frac\pi n+o(\frac1n)\right)=\pi+o(1)\to\pi$$

Note about the edit and its rollback
The edit by Nerarith was incorrect and unnecessary: $f(x)=o(g(x))$ iff $f(x)=\epsilon(x) \cdot g(x)$ with $\epsilon(x)\to0$. Not to be confused with $f(x)=O(g(x))$, iff there is a constant $M$ such that $|f(x)|\le M|g(x)|$ (in both cases, it's supposed to hold when $x\to x_0$ or $x\to\infty$).
Here I wrote the minimal information necessary to find the limit. We could write $\sin\frac{\pi}{n}=\frac{\pi}{n}+O(\frac{1}{n^3})$, but it's not mandatory. However, we can't write $\sin\frac{\pi}{n}=\frac{\pi}{n}+o(\frac{1}{n^3})$, as it's wrong.
A: For small values of $x$, $\sin x\sim x$. So for large values of $x$, $\sin(1/x)\sim 1/x$. So as $n$ increases, $n\sin(\pi/n) \sim n\cdot\pi/n = \pi$.
A: As often, the shortest is to use equivalents:
$\sin x\sim_0 x$, hence $\sin\dfrac\pi n\sim_\infty\dfrac\pi n$ and
$$n\sin\frac\pi n\sim_\infty n\dfrac\pi n=\pi.$$
A: I just want to point out this can be derived from Archimedes' Method for calculating the area of a (unit) circle (here we apply it for a polygon with $2n$ sides)

