lim(x^3-ax+b)/(x-1)=1,x趋近1,则常数a= ,b=

lim(x^3-ax+b)/(x-1)=1,x趋近1,则常数a= ,b=
lim(x^3-ax+b)/(x-1)=1,x趋近1,则常数a= ,b=

lim(x^3-ax+b)/(x-1)=1,x趋近1,则常数a= ,b=
首先分母的极限必需为0,否则极限不存在,矛盾.
因此 1^3-a*1+b=0
a=1+b
带入
lim (x^3-(1+b)x+b)/(x-1)
=lim[x^3--x-b(x-1)]/(x-1)
=lim[x(x^2-1)-b(x-1)]/(x-1)
=lim(x(x+1)-b)
=2-b=1
b=1,a=2