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OKay I tried using a L'Hopital's Rule (immediately failed), series expansion (wrote the first three terms and gave up on this method), Squeeze's Theorem (couldn't get a proper lower bound), and I am absolutely stumped

$$\lim_{n\to \infty} \frac{\tan(\pi/n)}{n\sin^2(2/n)}$$

According to Mathematica, it converges to $\frac{\pi}{4}$, but I have no idea how. any insight is greatly appreciated

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up vote 7 down vote accepted

Hint: Your limit is $$\lim_{x\to0}\frac{\tan\pi x}{\pi x}\cdot\left(\frac{ 2x}{\sin 2x}\right)^2\cdot\frac{\pi}{4}$$

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Why is your limit going to 0? I am going to infinity – Hawk Sep 23 '12 at 3:41
Note that $x=1/n$. So, when $n\to\infty$, $x\to0$. – Tapu Sep 23 '12 at 3:42
@Tapu Actually, it is $x\to 0^+$. – Pedro Tamaroff Sep 23 '12 at 3:42
@PeterTamaroff, Yes, thanks! – Tapu Sep 23 '12 at 3:44

Write it as

$$\frac{{\tan \frac{\pi }{n}}}{{n{{\sin }^2}\frac{2}{n}}} = \left( {\frac{n}{\pi }\tan \frac{\pi }{n}} \right)\frac{{\frac{2}{n}}}{{\sin \frac{2}{n}}}\frac{{\frac{2}{n}}}{{\sin \frac{2}{n}}}\frac{\pi }{4}$$

and recall $$\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1$$

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How did you get two $\frac{2}{n}$ on top? – Hawk Sep 23 '12 at 3:44
@jak Explot $\frac a a=1$ as many times as you want. Also $a=\frac 1 {a^{-1}}$, right? – Pedro Tamaroff Sep 23 '12 at 3:46
what motivated you to write out $\dfrac{n}{\pi}\dfrac{2}{n}\dfrac{2}{n}\dfrac{\pi}{4}$? How did you see this? – Hawk Sep 23 '12 at 3:50
@jak I see.Basically, you want to match up every sine (there's two of them) to a $2n^{-1}$, and match the tangent to one $\pi n^{-1}$. Note the tangent is just sine over cosine. The cosine goes to $1$ as $\pi/n\to0$, and we are left with the sine. All in all, we match things up and use $\sin x/x\to 1$. – Pedro Tamaroff Sep 23 '12 at 3:56

Almost anything you try should work. I am more comfortable near $0$, so would probably want to let $x=1/n$ and find the limit as $x\to 0^{+}$. So we want $$\lim_{x\to 0^{+}} \frac{x\tan(\pi x)}{\sin^2 2x}.\tag{$1$}$$

Now if we want to use Taylor series, all we need is the first terms of the Taylor expansion about $x=0$. For the top this is $\pi x^2$, and for the bottom it is $(2x)^2$.

But the approach by Peter Tamaroff is very good, and gets you the answer efficiently from basic ideas.

We can also use L'Hospital's Rule. It can be done mechanically, but we will need to apply the Rule twice.

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Moderately clever rearrangements of the quotient yield easy solutions, like Peter Tamaroff’s, but you actually can use l’Hospital’s rule if you rearrange the fraction properly and are sufficiently persistent:

$$\begin{align*} \lim_{n\to \infty} \frac{\tan(\pi/n)}{n\sin^2(2/n)}&=\lim_{n\to\infty}\frac{\frac1n\tan(\pi/n)}{\sin^2(2/n)}\\ &=\lim_{n\to\infty}\frac{-\frac{\pi}{n^3}\sec^2(\pi/n)-\frac1{n^2}\tan(\pi/n)}{-\frac4{n^2}\sin(2/n)\cos(2/n)}\\ &=\lim_{n\to\infty}\frac{\frac{\pi}n\sec^2(\pi/n)+\tan(\pi/n)}{4\sin(2/n)\cos(2/n)}\\ &=\frac{\pi}4\left(\lim_{n\to\infty}\frac{\sec^2(\pi/n)}{\cos(2/n)}\right)\left(\lim_{n\to\infty}\frac1{n\sin(2/n)}\right)+\lim_{n\to\infty}\frac{\tan(\pi/n)}{4\sin(2/n)\cos(2/n)}\\ &=\frac{\pi}4\lim_{n\to\infty}\frac{1/n}{\sin(2/n)}+\lim_{n\to\infty}\frac{\tan(\pi/n)}{2\sin(4/n)}\\ &=\frac{\pi}4\cdot\frac12\lim_{n\to\infty}\frac{2/n}{\sin(2/n)}+\lim_{n\to\infty}\frac{-\frac{\pi}{n^2}\sec^2(\pi/n)}{-\frac8{n^2}\cos(2/n)}\\ &=\frac{\pi}8+\frac{\pi}8\lim_{n\to\infty}\frac{\sec^2(\pi/n)}{\cos(2/n)}\\ &=\frac{\pi}8+\frac{\pi}8\\ &=\frac{\pi}4\;. \end{align*}$$

This approach is a bit easier if you substitute $x=1/n$ to make it

$$\lim_{x\to 0^+}\frac{x\tan\pi x}{\sin^22x}\;.$$

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