How can I solve the limit: $$\lim_{n \to \infty}(n^2 \sin^2(\frac\pi n))$$
I believe that the answer is $\pi^2$ as:
Area of regular polygon in terms of apothem = $\frac {na^2} {\tan(\frac\pi 2 - \frac\pi n)}$. Then taking the limit as n approaches infinity: $$\lim_{n \to \infty}(\frac {na^2} {\tan(\frac\pi 2 - \frac\pi n)})$$ which should represent the area of a circle $\pi a^2$.
Then by L'hopital's (as the limit becomes $\frac\infty \infty$) I got $$\lim_{n \to \infty}(\frac {n^2} {\pi\sec^2(\frac\pi 2 - \frac\pi n)})) = \pi$$ which then becomes $$(\frac 1 \pi)(\lim_{n \to \inf}(n^2\cos^2(\frac\pi 2 - \frac\pi n)) = \pi$$
By a trig identity this means $$\lim_{x \to \infty}(n^2\sin^2(\frac \pi n)) = \pi^2.$$
I've probably made a mistake somewhere so far, but how would I go about finding what the limit equals from this point (without using the area of the circle).
