# calc problem

• December 12th 2006, 09:01 PM
peprann3213
calc problem
How do I use calculus to PROVE that the lim as n goes to infinity(n*sin(pi/n)) = n ???
• December 13th 2006, 03:59 AM
ThePerfectHacker
Quote:

Originally Posted by peprann3213
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,
$n\left( (\pi/n)-\frac{(\pi/n)^3}{3!}+... \right)$
Distribution laws hold for infinite series,
$\pi -\frac{(\pi/n)^2}{3!}+...$
The limit is clearly zero for all the terms after $\pi$:
$\pi+0+0+..=\pi$
• December 13th 2006, 04:05 AM
ThePerfectHacker
Another way is the use of a famous limit.
The convergence of,
$\lim_{x\to \infty} x\sin (\pi/x)$
Is eqivalent tot,
$\lim_{x\to 0} \frac{\sin(\pi x)}{x}$
Thus,
$\lim_{x\to 0} \pi \frac{\sin(\pi x)}{\pi x}$
Thus, (limit composition rule)
$\pi$