# $\lim_{n \to \infty} 2^n\cos\left(\frac{\pi}{2^n}\right)\sin\left(\frac{\pi}{2^n}\right)$ (Without L'Hospital)

I'm trying to find $$\lim_{n \to \infty} 2^n\cos\left(\frac{\pi}{2^n}\right)\sin\left(\frac{\pi}{2^n}\right)$$ I think the answer is $\pi$, but I don't know how to find it. Could you please show me the shortcut?

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Hint: The limit you are considering is exactly the same as

$$\lim_{x \to \infty} x \cos{\left(\frac{\pi}{x}\right)}\sin{\left(\frac{\pi}{x}\right)}$$

or alternatively, by replacing $x$ with $1/x$:

$$\lim_{x \to 0^+} \frac{\cos{\left(\pi x\right)} \sin{\left(\pi x\right)}}{x}$$

and finally, replacing $x$ by $x/\pi$,

$$\lim_{x \to 0^+} \frac{\cos{x} \sin{x}}{x/\pi}$$

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Consider a regular $2^{n}-\,$gon inscribed in a circle of radius $1$. Then the area of each of the $2^{n}$ isosceles triangles is $$A_n=\cos\frac{\pi}{2^n}\sin\frac{\pi}{2^n}$$

Thus the total area is $$T_n=2^n\cos\frac{\pi}{2^n}\sin\frac{\pi}{2^n}$$

As $n$ grows large, this will approximate the area of the circle better and better, and since this area is $\pi$, we get $$\lim_{n\to\infty}2^n\cos\frac{\pi}{2^n}\sin\frac{\pi}{2^n}=\pi$$

Of course, this is less rigorous than other approaches, but I think it is charming.

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In essence, you could use this as a roundabout proof that $\sin x /x \to 1$.. – nbubis Aug 25 '13 at 6:20
+1 I can imagine it in my mind :) Nice thinking – Imane Fateh Aug 25 '13 at 11:58

The answer is $\pi$. Use the fundamental limit $\lim_{x\to 0}\frac{\sin x}{x}=1$.

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The proof depends on the tools you can use. Perhaps it is obvious that $$\lim_{n\to\infty}\cos\left(\frac{\pi}{2^n}\right)=1,$$ since $\frac{\pi}{2^n}\to 0$, and the cosine function is continuous, and $cos(0)=1$.

Perhaps you also know that $\lim_{h\to 0} \frac{\sin h}{h}=1$. If you do, then rewrite $2^n \sin\left(\frac{\pi}{2^n}\right)$ as: $$\pi \frac{\sin\left(\frac{\pi}{2^n}\right)}{\frac{\pi}{2^n}}.$$

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@T.Bongers: Yes, that's what I mean. Thank you. – André Nicolas Aug 25 '13 at 4:04

$$2^n\cos\left(\frac{\pi}{2^n}\right)\sin\left(\frac{\pi}{2^n}\right)=2^{n-1}\sin\left(\frac{\pi}{2^{n-1}}\right) \sim 2^{n-1}\frac{\pi}{2^{n-1}}=\pi,$$

since $\sin(x)\sim x$ as $x\sim 0$.

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Have you read Theo's answer, Mhenni? – Pedro Tamaroff Aug 25 '13 at 6:06