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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.

David K
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dfd
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2 Answers2

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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 $1$.

KM101
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If you replace $\frac\pi{2^m}$ with x, it is obvious that $x$ goes to zero when $m$ goes to infinity. So your limit is equivalent to:

$$\lim_{x \to 0}\frac{\sin x}{x}$$

...which is 1. If you learn to evaluate just one limit, this is the one :) You can easily find several elementary proofs on the web and I really recommend this one from MSE.

Saša
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