Question

若

lim

x→1

x2-6x+5

x2-1

=a，則a=

 

，

lim

n→∞

(

1

a

+

1

a2

+

1

a3

+…+

1

an

)=

 

．

Answer 1

分析：由題意知

lim

x→1

x2-6x+5

x2-1

=

lim

x→1

(x-1)(x-5)

(x+1)(x-1)

=

lim

x→1

x-5

x+1

=-2，由此可知a=-2．所以

lim

n→∞

(

1

a

+

1

a2

+

1

a3

+…+

1

an

)=

lim

n→∞

1 -2 [1-(- 1 2 )n]

1- (- 1 2 )

，進而可得答案．

解答：解：∵

lim

x→1

x2-6x+5

x2-1

=

lim

x→1

(x-1)(x-5)

(x+1)(x-1)

=

lim

x→1

x-5

x+1

=-2，
∴由

lim

x→1

x2-6x+5

x2-1

=a，知a=-2．
∴

lim

n→∞

(

1

a

+

1

a2

+

1

a3

+…+

1

an

)=

lim

n→∞

1 -2 [1-(- 1 2 )n]

1- (- 1 2 )

=-

1

3

．
答案：-2，-

1

3

．

點評：本題考查數列的極限，解題的關鍵是合理轉化，消除零因子．