请问这个极限怎么求n求极限lim(n->∞）∑sin(k/n^2),其中n^2代表n的平方k=1

u(n) = ∑sin(k/n^2) = sin(1/n^2) +sin(2/n^2) + .+ sin(n/n^2)
sin(1/n^2) + sin(n/n^2) = 2 sin [(n+1)/(2n^2)] cos [(n-1)/(2n^2)] 和差化积
sin(2/n^2) + sin[(n-1)/n^2] = 2 sin [(n+1)/(2n^2)] cos [(n-3)/(2n^2)]
……
当n为偶数时,
u(n) = 2 sin [(n+1)/(2n^2)] { cos [(n-1)/(2n^2)] + cos [(n-3)/(2n^2)] + …… + cos [1/(2n^2)] }
=> n * sin [(n+1)/(2n^2)] cos [(n-1)/(2n^2)] ≤ u(n)≤ n * sin [(n+1)/(2n^2)]
当n->∞时,n * sin [(n+1)/(2n^2)] cos [(n-1)/(2n^2)] -> 1/2
n * sin [(n+1)/(2n^2)] -> 1/2
=> 当n->∞时,u(n) -> 1/2
当n为奇数时,类似可求.
Lim ∑sin(k/n^2) = 1/2

2. (x趋向于0) =limx^3/(x-sinx) =lim3x^2/(1-cosx) =lim6x/sinx[ln(lnx)]

化成积分来做