设lim(x→∞)（(x^3+x^2+1)^1/3-ax-b）=0求常数a,b

Question

设lim(x→∞)（(x^3+x^2+1)^1/3-ax-b）=0求常数a,b

Answer 1

a=lim(x->∞) [(x^3+x^2+1)^(1/3)]/x
=lim(x->∞) [(1+1/x+1/x^3)^(1/3)
=1
b=lim(x->∞) [(x^3+x^2+1)^(1/3)-x]
=lim(x->∞) [(x^3+x^2+1-x^3]/[(x^3+x^2+1)^(2/3)+x(x^3+x^2+1)^(1/3)+x^2]
=lim(x->∞) [x^2+1]/[(x^3+x^2+1)^(2/3)+x(x^3+x^2+1)^(1/3)+x^2]
==lim(x->∞) [1+1/x^2]/[(1+1/x+1/x^3)^(2/3)+(1+1/x+1/x^3)^(1/3)+1]
=1/(1+1+1)
=1/3