设limx→-1 (x^3+ax^2-x+4)/x+1=b(b为非零常数),求a,b

问题描述:

设limx→-1 (x^3+ax^2-x+4)/x+1=b(b为非零常数),求a,b

lim(x→-1 )(x^3+ax^2-x+4)/(x+1)=b
所以x=-1分子x^3+ax^2-x+4=0
代入得
-1+a+1+4=0
a=-4
lim(x→-1 )(x^3+ax^2-x+4)/(x+1) (0/0)
=lim(x→-1 ) 3x^2+2ax-1
=3-2a-1
=10=b因为分母等于0,如果分子不等于0,那么极限就不存在