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At lunch a coworker was talking about how to calculate, say, the 100th digit of pi using a square around the circle, then a pentagon, etc, basically you end up taking the limit of the circumference as the number of sides n goes to infinity.

So I tried working out the math, but I got stuck at proving:

$$\lim_{n \to \infty} 2n\tan\frac{\pi}{n} = 2 \pi$$

Any ideas how?

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5 Answers 5

up vote 4 down vote accepted

Putting $n=\frac1h, h\to0$ as $n\to\infty$

$$\lim_{n \to \infty} 2n\cdot\tan\frac{\pi}{n}$$

$$=2\lim_{h\to0}\frac{\tan \pi h}h$$

$$=2\pi\lim_{h\to0}\frac{\sin \pi h}{\pi h}\cdot \frac1{\lim_{h\to0}\cos\pi h}$$

We know, $\lim_{x\to0}\frac{\sin x}x=1$ and $\lim_{x\to0}\cos x=1$

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Ahh makes sense thank you! –  user1956609 Jul 10 '13 at 18:21
    
@user1956609, my pleasure. Conversion of the limit $0$ by proper replacement has made things easier me in many cases –  lab bhattacharjee Jul 10 '13 at 18:23

From Taylor series we know that $$\tan x=_0x +o(x)$$ and if $y\to+\infty$ then $\frac{x}{y}\to 0$ hence we have $$\tan\left(\frac{x}{y}\right)y=_\infty\left(\frac{x}{y}+o\left(\frac{x}{y}\right)\right)y=_\infty x+o(1)$$ so we can conclude.

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Hope you're having a great day! –  amWhy May 21 at 11:43

As shown in this answer, $$ \lim_{x\to0}\frac{\tan(x)}{x}=1\tag{1} $$ For $x\ne0$, $(1)$ is equivalent to $$ \lim_{y\to\infty}\frac{\tan(x/y)}{x/y}=1\tag{2} $$ multiplying $(2)$ by $x$ yields $$ \lim_{y\to\infty}y\tan(x/y)=x\tag{3} $$ The case $x=0$ is verified trivially.

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Hint: Do you know that $\lim_{x\to 0} \frac{\sin x}{x} = 1$?

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$\lim_{n \to \infty} 2n(tan\frac{\pi}{n}) = \lim_{n \to \infty} 2\pi \frac{\tan \frac{\pi}{n}}{\frac{\pi}{n}}=\lim_{x\to 0}2\pi \frac{\tan x}{x}= 2\pi$

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