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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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An idea:

$$\frac{\tan\frac\pi{n} }{n\sin^2\frac2n}=\left(\frac{\frac2n}{\sin\frac2n}\right)^2\cdot\frac{\sin\frac\pi n}{\frac\pi n}\cdot\frac1{\cos\frac\pi n}\cdot\frac\pi4$$

And now use that

$$\frac{\sin\left( f(n)\right)}{f(n)}\xrightarrow[n\to\infty]{}1\;\;\text{whenever}\;\;\lim_{n\to\infty}f(n)=0$$

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1  
I typed wrong and cannot delete a stupid comment. – Claude Leibovici Jan 31 '14 at 7:20

The inequalities you get from the unit circle are, for $0\leq t<\pi/2$, $$ \sin t \leq t \leq \tan t. $$ In particular, as $\tan t=\sin t/\cos t$, we have $\sin t\geq t\cos t$. Now $$ \dfrac{\tan\left(\dfrac{\pi}{n}\right)}{n\sin^2\left(\dfrac{2}{n}\right)}\geq\dfrac{\dfrac\pi n}{n\,\dfrac{4}{n^2}}=\frac\pi4. $$ And $$ \dfrac{\tan\left(\dfrac{\pi}{n}\right)}{n\sin^2\left(\dfrac{2}{n}\right)} =\dfrac1{\cos\dfrac\pi n}\,\dfrac{\sin\left(\dfrac{\pi}{n}\right)}{n\sin^2\left(\dfrac{2}{n}\right)}\leq\dfrac1{\cos\dfrac\pi n}\,\dfrac{\dfrac\pi n}{n\,\dfrac4{n^2}\,\cos^2\dfrac2n}=\dfrac\pi4\,\dfrac{1}{\cos\dfrac\pi n\,\cos^2\dfrac2n}\to\dfrac\pi4. $$

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