How do I use calculus to PROVE that the lim as n goes to infinity(n*sin(pi/n)) = n ???
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How do I use calculus to PROVE that the lim as n goes to infinity(n*sin(pi/n)) = n ???
Express it an a Taylor series centered at zero,Quote:
Originally Posted by peprann3213
$\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 $\displaystyle \pi$:
$\displaystyle \pi+0+0+..=\pi$
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)
$\displaystyle \pi$
