Question

（2008•靜安區(qū)一模）計算：

lim

n→∞

(2n-

4n2+2n-1

2n+2

)=

1

．

Answer 1

分析：由

lim

n→∞

(2n-

4n2+2n-1

2n+2

)=

lim

n→∞

4n2+4n-4n2-2n+1

2n+2

=

lim

n→∞

2n+1

2n+2

可求

解答：解：∵

lim

n→∞

(2n-

4n2+2n-1

2n+2

)=

lim

n→∞

4n2+4n-4n2-2n+1

2n+2

=

lim

n→∞

2n+1

2n+2

=

lim

n→∞

1+ 1 2n

1+ 1 n

=1
故答案為：1

點評：本題主要考查了

∞

∞

型的數(shù)列極限的求解，屬于基礎(chǔ)試題