lim(x->∞)[(a+1)x^3+bx^2+2] /[2x^2+x+1 ]=-2
a+1 =0
=> a=-1
and
b/2 = -2
b= -4能否再详细些？能否再详细些？分母 =x^3
分子 =x^2
lim(x->∞) f(x) ->∞
所以分母 只能是x^2
=> a-1 =0

lim(x->∞)[bx^2+2] /[2x^2+x+1 ]=-2

lim(x->∞)[b+2/x^2] /[2+1/x+1/1/x^2 ]=-2
b/2 =-2
b=-4