求极限lim(n*sin(pi/n)) (n->无穷大)

lim(n*sin(pi/n)) (n->无穷大)
=lim[sin(pi/n)/(1/n)] (n->无穷大)
=pi*lim[sin(pi/n)/(pi/n)] (n->无穷大)
令pi/n=x [n->无穷大时,x->0]
原式可化为
pi*lim[sinx/x] (x->0)
=pi*lim(cosx/1)(x->0)[根据洛必塔法则]
=pi