Numerical calculation of the limit of the area of regular n-sided polygon as n goes to infinity.

Question

Consider $n$-sided regular polygon. Then we can calculate the area by splitting it into isosceles triangles and using the formula for triangles: $A=\frac{1}{2}ab\sin\gamma$, then we can write the area of an $n$-sided polygon as: $$A_n=n\frac{1}{2}r^2\sin{\frac{2\pi}{n}}$$ where $r$ is the radius of the excircle of the polygon. Now when we take this as $n\to\infty$ this should give us the area of the circle, which is $\pi r^2$. How does one show numerically that $$\lim_{n\to\infty}{\frac{1}{2}n\sin{\frac{2\pi}{n}}}$$ equals $\pi$. (Of course I am interested in the way without L'Hospital's rule)

Answer 1

$$\begin{align}\lim_{n\to\infty}{\frac{1}{2}n\sin{\frac{2\pi}{n}}}&=\lim_{n\to\infty}\pi\cdot \frac{n}{2\pi} \sin\frac{2\pi}n\\ &=\lim_{t \to 0} \pi \cdot\frac{\sin t}{t}=\pi\end{align}$$ by substituting $t = \frac{2\pi}{n}$.