Question

给出下面四个式子：
①

lim

x→2

x2-4

x-2

②

lim

x→∞

4x2+3

6x2-5

③

lim

x→+∞

(

x2+2

-

x2-2

)
④

lim

x→2

3 x-1 -1

x-1 -1

；
其中极限等于

2

3

的有（　　）

A．0个

B．1个

C．2个

D．3个

Answer 1

分析：由①

lim

x→2

x2-4

x-2

=

lim

x→2

(x+2)②

lim

x→∞

4x2+3

6x2-5

=

lim

x→∞

4+ 3 x2

6- 5 x2

③

lim

x→+∞

(

x2+2

-

x2-2

)=

lim

x→+∞

4

2+x2+ x2-2

④x-1=t⁶，则

lim

x→2

3 x-1 -1

x-1 -1

=

lim

t→1

t2-1

t3-1

=

lim

t→1

t+1

t2+t+1

，代入即可求解

解答：解：∵①

lim

x→2

x2-4

x-2

=

lim

x→2

(x+2)=4
②

lim

x→∞

4x2+3

6x2-5

=

lim

x→∞

4+ 3 x2

6- 5 x2

=

2

3

③

lim

x→+∞

(

x2+2

-

x2-2

)=

lim

x→+∞

4

2+x2+ x2-2

=0
④x-1=t⁶，则

lim

x→2

3 x-1 -1

x-1 -1

=

lim

t→1

t2-1

t3-1

=

lim

t→1

t+1

t2+t+1

=

2

3

故选C

点评：本题主要考查了函数极限的求解方法与技巧，属于基础试题