I was able to compute the limit of the following using l'Hopital's rule, and found it to be $\frac{\pi}{4}$, but apparently there is a way to evaluate the limit using the squeeze theorem, apparently you have the fact that the area of a sector of a circle with radius 1 squeezes between the area of a larger triangle and smaller one for small angles
Here is the limit:
$\displaystyle \lim_{n \rightarrow \infty} \dfrac{\tan\left(\dfrac{\pi}{n}\right)}{n\sin^2\left(\dfrac{2}{n}\right)}$
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