I would like to calculate $$\lim_{n\rightarrow \infty}\frac{n}{4} \sin \left(\frac{4 \pi}{n} \right)$$
Clearly this is a limit of the type $\infty \cdot 0$, so I'm thinking there is probably some way to turn it to $\infty / \infty$ or $0 / 0 $ and then use L'Hopital but I can't think of any such trick. I cannot think of a way to do it without L'Hoptial either. Thanks for any input.
HINT: If you can prove that $$ \lim_{x\to 0}\frac{\sin x}{x}=1, $$ then you are almost done by writing $x=4/n$ and letting $n\to \infty$.
HINT: Write your expression as $$\lim_{n\to\infty} \dfrac{\sin(\pi(4/n))}{4/n}$$ and make use of the fact that $$\lim_{x \to 0} \dfrac{\sin(ax)}{x} = a$$
you can use the taylor of $\sin(4\pi/n)$ as follow $$y=x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots$$ $$y=\frac{4\pi}{n}-\frac{(4\pi)^3}{3!n^3}+\frac{(4\pi)^5}{5!n^5}+\cdots$$ so $$\lim_{n\rightarrow \infty }\frac{n}{4}(\frac{4\pi}{n}-\frac{(4\pi)^3}{3!n^3}+\frac{(4\pi)^5}{5!n^5}+\cdots)=\pi$$
We know $\sin \theta = \theta + \mathcal{O}(\theta^3)$ for $\theta \approx 0$. Then it becomes $\frac{n}{4}\frac{4\pi}{n} + \mathcal{O}(n^{-2})$.
O notation can be handy.
