Question

高数极限怎么求未知数？

Answer 1

5.
(1)
lim(x->1) (x^2+ax+b)/(1-x) = 5 (0/0)
1^2+a(1) +b =0
a+b =-1 (1)
lim(x->1) (x^2+ax+b)/(1-x) = 5 (0/0 分子分母分别求导)
lim(x->1) -(2x+a) = 5
-(2+a)=5
a=-7
from (1)
a+b =-1
-7+b=-1
b=6
(a,b)=(-7,6)
(2)
4x^2 +3
=4x(x-1) +4x+3
=4x(x-1) +4(x-1) +7
/
lim(x->∞) [ (4x^2+3)/(x-1) +ax+b ] = 0

lim(x->∞) [ 4x+4 + 7/(x-1) +ax+b ] = 0
lim(x->∞) [ (4+a)x+(4+b) + 7/(x-1) ] = 0
x的系数=0
4+a=0
a=-4
lim(x->∞) [ (4+a)x+(4+b) + 7/(x-1) ] = 0
lim(x->∞) [ (4+b) + 7/(x-1) ] = 0
4+b=0
b=-4
(a,b)=(-4,-4)