Step by step explanation of this example usage of L'Hôpital's rule? Using L'Hôpital's rule, I need to show how:
$$\lim_{n\to\infty}\frac{p^2\cot(\frac{\pi}{n})}{4n}=\pi r^2$$
Where $p$ is the perimiter of a regular polygon and $r$ is a radius. The idea of the example is to prove that an infinitely sided polygon becomes a circle. 
The perimeter is fixed. The formula is actually for calculating the area of a regular polygon from its perimeter and the number of sides- therefore once $n$ reaches infinity, $p$ will become a circumference.
 A: This is an infinity/infinity form:
$$
\begin{align}
   \lim_{n\rightarrow\infty} \frac{p^{2}\cot(\pi/n)}{4n}
       & = \frac{p^{2}}{4\pi}\lim_{n\rightarrow\infty}\cos(\pi/n)\frac{(\pi/n)}{\sin(\pi/n)}
 \\
       & = \frac{p^{2}}{4\pi}\lim_{n\rightarrow\infty}\cos(\pi/n)
           \lim_{n\rightarrow\infty}\frac{(\pi/n)}{\sin(\pi/n)} \\
       & = \frac{p^{2}}{4\pi}\lim_{l\rightarrow 0}\frac{l}{\sin(l)}=\frac{p^{2}}{4\pi}\lim_{l\rightarrow 0}\frac{1}{\cos(l)}=\frac{p^{2}}{4\pi}
\end{align}
$$
A: With $m = \pi/n$, we easily find $$\lim_{n \to \infty} \frac{\cot (\pi/n)}{n} = \lim_{m \to 0^+} \frac{m}{\pi} \cot m = \frac{1}{\pi} \lim_{m \to 0^+} \frac{m}{\tan m} = \frac{1}{\pi} \lim_{m \to 0^+} \cos m \cdot \frac{m}{\sin m}.$$  L'Hopital's rule applied to the fraction $m/\sin m$ immediately yields $1/\pi$ as the value of the limit.
A: $\cot (\frac{\pi}{n}) \to_n \cot 0$, so you can expand it around $0: \cot \frac{\pi}{n} \sim \frac{n}{\pi} - \frac{\pi}{3 n} + O(n^{-3})$. Once you multiply the terms of this Maclaurin expansion by the remaining terms (I understand $p = (2 \pi r)^2$) you get the result. 
This is not exactly L'Hospital's rule, but quite close. 
