How do I use calculus to PROVE that the lim as n goes to infinity(n*sin(pi/n)) = n ???

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- Dec 12th 2006, 09:01 PMpeprann3213calc problem
How do I use calculus to PROVE that the lim as n goes to infinity(n*sin(pi/n)) = n ???

- Dec 13th 2006, 03:59 AMThePerfectHacker
Express it an a Taylor series centered at zero,

$\displaystyle n\left( (\pi/n)-\frac{(\pi/n)^3}{3!}+... \right) $

Distribution laws hold for infinite series,

$\displaystyle \pi -\frac{(\pi/n)^2}{3!}+...$

The limit is clearly zero for all the terms after $\displaystyle \pi$:

$\displaystyle \pi+0+0+..=\pi$ - Dec 13th 2006, 04:05 AMThePerfectHacker
Another way is the use of a famous limit.

The convergence of,

$\displaystyle \lim_{x\to \infty} x\sin (\pi/x)$

Is eqivalent tot,

$\displaystyle \lim_{x\to 0} \frac{\sin(\pi x)}{x}$

Thus,

$\displaystyle \lim_{x\to 0} \pi \frac{\sin(\pi x)}{\pi x}$

Thus, (limit composition rule)

$\displaystyle \pi$