已知函数f(x)=
|
(2x+3)=lim x→0+
(2x+3)=3,lim x→0-
f(0)=a点在x=0处连续,
所以
f(x)=f(0),lim x→0
即a=3,
故lim x→∞
=3n2+1 32n2+n
=3 9
.1 3
故答案为:
.1 3
已知函数f(x)=
|
(2x+3)=lim x→0+
(2x+3)=3,lim x→0-
f(0)=a点在x=0处连续,
所以
f(x)=f(0),lim x→0
即a=3,
故lim x→∞
=3n2+1 32n2+n
=3 9
.1 3
故答案为:
.1 3