例1求lm( 十∴十 n→0 n2+1√n2+2 2 n+n 解 十∴ < n2+n√n2+1 n+n n-+ 1 又lm = n→√n2+n n→0 由夹逼定理得 n2+1 十 十 十∴十 n2+1√n2+2 n+n例1 ). 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 