Question

设lim(x→-1) x³-ax²-x+4/x+1具有极限l,试求a和l的值

Answer 1

∵lim(x→-1)（ x³-ax²-x+4）/（x+1）=l
∴x^3-ax^2-x+4能被x+1整除。
∴（ x³-ax²-x+4）=(x+1)[x²-(a+1)x+4]
∴-(a+1)x+4x=-x
∴3-a=-1,a=4
∴lim(x→-1)（ x³-ax²-x+4）/（x+1）
=lim(x-->-1)(x^2-5x+4)(x+1)/(x+1)
=lim(x-->-1)(x^2-5x+4)
=10
∴a=4,l=10

Answer 2

x→-1 极限=l
x³-ax²-x+4可以分解含有x+1因式
x³-ax²-x+4=(x^3-x)-(ax^2-4)=x(x^2-1)-(ax^2-4)
所以x=-1 ax^2-4=0 a=4
x³-ax²-x+4=x(x-1)(x+1)-4(x-1)(x+1)
=(x+1)(x^2-x-4x+4)
=(x+1)(x^2-3x+4)
lim(x→-1) x³-ax²-x+4/x+1
=lim(x→-1) (x^2-3x+4)=8
l=8

Answer 3

x^3-ax^2-x+4能被x+1整除。
x^3-ax^2-x+4=(x^3+x^2)-(a+1)(x^2+x)+a(x+1)-a+4。
所以，-a+4=0，即a=4。
x^3-ax^2-x+4=x^3-4x^2-x+4=x(x^2-1)-4(x^2-1)=(x^2-1)(x-4)
(x^3-ax^2-x+4)/(x+1)=(x^2-1)(x-4)/(x+1)=(x-1)(x-4)
l=lim(x-->-1)(x^3-ax^2-x+4)/(x+1)=lim(x-->-1)(x-1)(x-4)=-2*(-5)=10

Answer 4

lim(x→-1) (x³-ax²-x+4)/(x+1)) (0/0)
x³-ax²-x+4=0 (at x=-1)
-1-a+1-4=0
a=-4
lim(x→-1) (x³-ax²-x+4)/(x+1))
=lim(x→-1) (x³+4x²-x+4)/(x+1)) (0/0)
=lim(x→-1) (3x^2+8x-1)/1
=3-8-1
=-6