问题
填空题
|
答案
∵lim x→1
=1,
+bx+a x2-1
∴lim x→1
=
+bx+a x2-1 lim x→1
=1x+a-b2 (x+1)(x-1)(
-b)x+a
∴
,a-b2=-1
=11 2(
-b)a+1
a=-
,b=-15 16
.1 4
故答案为:-
;-15 16
.1 4
|
∵lim x→1
=1,
+bx+a x2-1
∴lim x→1
=
+bx+a x2-1 lim x→1
=1x+a-b2 (x+1)(x-1)(
-b)x+a
∴
,a-b2=-1
=11 2(
-b)a+1
a=-
,b=-15 16
.1 4
故答案为:-
;-15 16
.1 4