Why do $4\cdot 2^n\sin\frac{45}{2^n}$, $2\cdot 2^n\sin\frac{90}{2^n}$, and $1\cdot 2^n\sin\frac{180}{2^n}$ all tend to $\pi$?

Question

I am not sure what question or inquiries to ask actually, but I just think this is really awesome

Can someone explain to me why the graphs of $$4\cdot 2^n\sin\frac{45}{2^n}, \qquad 2\cdot 2^n\sin\frac{90}{2^n}, \qquad\text{and}\qquad 1\cdot 2^n\sin\frac{180}{2^n}$$ all tend to $\pi$?

Answer 1

Because $\sin x \approx x$ when $x$ is small and measured in radians. When $x$ is measured in degrees $\sin x \approx \frac {\pi x}{180}.$ When $n$ gets large the argument of $\sin$ becomes small.

Answer 2

This is due to $\lim_{h \to 0}\frac{\sin h}h = 1$.

Hence \begin{align}\lim_{n \to \infty}4 \cdot 2^n \cdot \sin\left(\frac{45^\circ}{2^n}\right)&=\lim_{n \to \infty}4 \cdot 2^n \cdot \sin\left(\frac{\pi}{2^{n+2}}\right) \\ &=\lim_{n \to \infty}4 \cdot 2^n \cdot \frac{\pi}{2^{n+2}}\cdot \frac{\sin\left(\frac{\pi}{2^{n+2}}\right)}{\frac{\pi}{2^{n+2}}} \\ &=\lim_{n \to \infty}4 \cdot 2^n \cdot \frac{\pi}{2^{n+2}}\cdot \lim_{n \to \infty}\frac{\sin\left(\frac{\pi}{2^{n+2}}\right)}{\frac{\pi}{2^{n+2}}} \\ &= \pi \cdot 1\\ &= \pi\end{align}

Similarly for the other sequences.

That is we have

$$\lim_{n \to \infty}w \cdot 2^n \cdot \sin \left( \frac{180^\circ}{w\cdot 2^n}\right)=\pi$$