求极限lim(x趋于1)(x^3-ax^2-x+4)/(x+1)=b a b为实数
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求极限lim(x趋于1)(x^3-ax^2-x+4)/(x+1)=b a b为实数
求极限lim(x趋于1)(x^3-ax^2-x+4)/(x+1)=b
a b为实数
求极限lim(x趋于1)(x^3-ax^2-x+4)/(x+1)=b a b为实数
应该是趋于-1吧?
如果是的话,分母上x^3-ax^2-x+4应该含(x+1)因式
x^3+x^2-(a+1)x^2-(a+1)x+(a+2)x+(a+2)+4-(a+2)
=(x+1)(x^2-(a+1)x+(a+2))+2-a
极限lim(x趋于-1)(x^3-ax^2-x+4)/(x+1)
=lim(x趋于-1)((x+1)(x^2-(a+1)x+(a+2))+2-a)/(x+1)
=lim(x趋于-1)(x^2-(a+1)x+(a+2))+(2-a)/(x+1)
=(-1)^2+(a+1)+(a+2)+lim(x趋于1)(2-a)/(x+1)
=2a+6+lim(x趋于1)((2-a)/(x+1)
=b
则有:2-a=0
a=2
b=2a+6=10
a=2时b=-2;
a不等于2时,b=(4-a)/2
(4-a)/2
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