例3求lm( ∴十 n2+1√m2+2 √n-+n n 解 < 十∴ n tn n2+1 √n2+nn2+1 又l m =lim n→0√n2+nn→ 1+ l,由夹逼定理得 n→√n2+1n→0 ∴十 n→0 √n2+1√n2+2 2 √n-+n例3 ). 1 2 1 1 1 lim( 2 2 2 n n n n + n + + + + → + 求  解 , 1 1 1 1 2 2 2 2 +  + + + +  + n n n n n n n n   n n n n n n 1 1 1 lim 2 lim + = → + → 又 = 1, 2 2 1 1 1 lim 1 lim n n n n n + = → + → = 1, 由夹逼定理得 ) 1. 1 2 1 1 1 lim( 2 2 2 = + + + + + n→ n + n n n 