I am currently working on a proof with a good friend of mine that involves adding more and more triangles to the sides of a regular polygon but keeping the longest diagonal constant until eventually, it becomes a circle. And we ended up with this formula.
$4$-sided regular→$8$-sided regular→$16$-sided regular→$32$-sided regular$\to \ldots \to n$-sided regular
(When $n$ tends to infinity, the area will be equal to that of a circle with the longest diagonal as diameter)
The Question is:
Can someone explain in detail:
Why does $$\frac {\sin \frac {\pi}{2^m}}{\frac{\pi}{2^m}} = 1$$ when $m$ tends to infinity?
Thanks in advance :D
Note I am still a beginner in finding limits. It would help greatly if you can explain step by step.
Because we have
$$\lim_{x \to 0}\frac{\sin x}{x} = 1$$
which is a standard limit typically proven through simple geometrical observations followed by the Squeeze Theorem. You can check this link for a various explanation of why this works, particularly the one by robjohn, which I find the simplest and most elegant.
Another way of stating the above limit is
$$\sin x \sim x; \quad x \to 0$$
(Obviously, if $\frac{\sin x}{x} \to 1$ as $x \to 0$, then $\sin x$ becomes very close to $x$.)
In case of your limit, you have
$$\lim_{m \to \infty} \frac{\sin\left(\frac{\pi}{2^m}\right)}{\frac{\pi}{2^m}}$$
It’s clear that as $m \to \infty$, $\frac{\pi}{2^m} \to 0$ due to the growing denominator, so we make a substitution $t = \frac{\pi}{2^m}$. Clearly, $t \to 0$ as $m \to \infty$, so the limit becomes
$$\lim_{t \to 0}\frac{\sin t}{t}$$
which becomes $1$.