n(n+1) (2)因为 n n2+nn2+1 n(n+1) -n(nt -n(n+ 1) 月→n+n lim (3)因为 ≤ √n2+1 又 lim nsin lim-vn2 =π,同理 lim nsir lim(sin +sin +…+Sin n2+1 n2+2 (4)当x>0时,1<√1+x<1+x,故lmh+x=1 当-1<x<0时,1+x<h+x<1,故imy1+x=1 故 lim√+x=1 4.利用单调有界准则证明下面数列存在极限,并求其极限值: √2…,a=√2(m次复合 (2)x=1,x2=1+-x 高十1”=1+如 证(1)易知an=√2an(m=12…),下证此数列单调有界 √2<2,假设n=k时 2a<2,即an<2(n=12…),即此数列有界3 (2) 因为 2 2 2 2 22 2 1 ( 1) 2 12 12 1 2 n n n n n n n nn n n nn n n n + = + ++ < + ++ + + + + ++ + " " 22 2 2 1 ( 1) 1 2 2 11 1 1 n n n nn n n + < + ++ = ++ + + " , 又 2 2 1 1 ( 1) ( 1) 2 2 1 lim lim n n 1 2 nn nn →∞ →∞ nn n + + = = + + , 故 22 2 12 1 lim ( ) n 1 2 2 n →∞ n n nn + ++ = ++ + " . (3) 因为 2 22 2 2 πππ π π sin sin sin sin sin 12 1 n n nn n n nn n ≤ + ++ ≤ + ++ + + " , 又 2 2 2 2 π sin π π lim sin lim π n n π n n n n nn nn n n →∞ →∞ + = ⋅= + + + , 同理 2 π lim sin π 1 n n n →∞ = + , 故 22 2 ππ π lim (sin sin sin ) π 1 2 n n n nn →∞ + ++ = ++ + " . (4) 当 x > 0 时, 11 1 n < + <+ x x , 故 0 lim 1 1 n x x → + + = ; 当 −< < 1 0 x 时, 1 11 n +< +< x x , 故 0 lim 1 1 n x x → − + = . 故 0 lim 1 1 n x x → + = . 4. 利用单调有界准则证明下面数列存在极限, 并求其极限值: (1) 1 2 2, 2 2, , 2 2 2 n aa a == = " " ( n 次复合); (2) 1 1 1 2 1 1 1, 1 , , 1 1 1 n n n x x xx x x x − − = =+ =+ + + " . 证 (1) 易知 1 2 ( 1,2, ) n n a an + = = " , 下证此数列单调有界: 当 n =1时, 1 a = < 2 2 , 假设 n k = 时, 2 k a < , 则当 n k = +1时, k 1 a + = 2 k a < 2 , 即 2( 1,2, ) n a n < = " , 即此数列有界;