Question

数学极限问题

数学极限问题x->1 x^2+ax+b/x^2-1=3，则a，b取何值？

Answer 1

lim(x->1) (x^2+ax+b)/(x^2-1) = 3 (0/0)
1^2+a(1) +b = 0
a+b= -1 (1)
lim(x->1) (x^2+ax+b)/(x^2-1) = 3 (0/0)
lim(x->1) (2x+a)/(2x) = 3
(2+a)/2 =3
2+a=6
a=4
from (1)
4+b=-1
b= -5
(a,b)= (4,-5)