$\lim_{n \to \infty} 2^n\cos\left(\frac{\pi}{2^n}\right)\sin\left(\frac{\pi}{2^n}\right)$ (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?
 A: 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 $x/\pi$,
$$\lim_{x \to 0^+} \frac{\cos{x} \sin{x}}{x/\pi}$$
A: 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.
A: The answer is $\pi$. Use the fundamental limit $\lim_{x\to 0}\frac{\sin x}{x}=1$.
A: 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:
$$\pi \frac{\sin\left(\frac{\pi}{2^n}\right)}{\frac{\pi}{2^n}}.$$
A: $$ 2^n\cos\left(\frac{\pi}{2^n}\right)\sin\left(\frac{\pi}{2^n}\right)=2^{n-1}\sin\left(\frac{\pi}{2^{n-1}}\right) \sim 2^{n-1}\frac{\pi}{2^{n-1}}=\pi, $$
since $\sin(x)\sim x$ as $x\sim 0$.
