求上下极限lim(x趋近无穷大){∫(o到x) sintdt}/x

|sint|是周期为π的函数 ∫[0->π] |sint|dt = 2 则有∫[0->nπ] |sint|dt = 2n,∫[0->(n+1)π] |sint|dt = 2(n+1) 由于f(x)=∫[0->x] |sint|dt是个增函数，所以当x∈[nπ, (n+1)π]时,f(nπ)<=f(x)<=f((n+1)π), 易得:f(nπ)/((n+1)π)<=f(x)/x<=f((n+1)π)/nπ 同时取极限:x->∞时，n->∞，lim[n->∞] f(nπ)/((n+1)π) = lim[n->∞] 2n/((n+1)π) = 2/π lim[n->∞] f((n+1)π)/nπ 亦等于2/π， 由夹逼定理得lim[x->∞] ∫[0->x] |sint|dt / x =2/π