极限x趋向于1求(x-1)*(x+3)⼀x^3-2x^2+ax+b=-2⼀3中a,b的值

2024-11-16 17:35:12
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回答1:

x趋向于1求(x-1)(x+3)/(x^3-2x^2+ax+b)= -2/3
分子趋于0,但极限不为0,故分母趋于0,即:a+b-1=0
x^3-2x^2+ax+b=x^3-2x^2+ax+1-a=x^3-x^2-x^2+x-x+ax+1-a=(x-1)(x^2-x+a-1)
所以:lim(x-1)(x+3)/(x^3-2x^2+ax+b)=lim(x+3)/(x^2-x+a-1)=4/(a-1)
由4/(a-1)=-2/3得: a=-5,从而b=6