Question

lim

n→∞

[

n2+an

-(bn+1)]=b，則a的值是（　�。�

A．4

B．2

C．-2

D．不確定

Answer 1

分析：通過(guò)分子有理化，然后分子與分母同除n，利用數(shù)列極限的運(yùn)算法則，推出關(guān)系式，求出b，a的值即可．

解答：解：

lim

n→∞

[

n2+an

-(bn+1)]
=

lim

n→∞

[ n2+an -(bn+1)]•[ n2+an +(bn+1)]

n2+an +(bn+1)

=

lim

n→∞

n2+an-(bn+1)2

n2+an +(bn+1)

=

lim

n→∞

n2+an-b2n2-2bn-1

n2+an +(bn+1)

=

lim

n→∞

n2(1-b2)+an-2bn-1

n2+an +(bn+1)

=

lim

n→∞

n(1-b2)+a-2b- 1 n

1+ a n +(b+ 1 n )

=b
∴1-b²=0，解得b=1或b=-1（舍去），且

a-2b

1+b

=b，
∴a=4．
故選：A．

點(diǎn)評(píng)：本題考查數(shù)列的極限的運(yùn)算法則，分子有理化的運(yùn)算，考查計(jì)算能力．