We have `underset(xto0)lim((4x-1)^(1/3)+a+bx)/(x)=1/3`
`implies" "underset(xto0)lim(-[1-4/3x]+a+bx)/(x)=1/3`
`implies" "underset(xto0)lim((a-1)+(4/3+b)x)/(x)=1/3`
For limit to exist, `a-1=0` or a = 1
`implies" "underset(xto0)lim((4/3+b)x)/(x)=1/3`
`4/3+b=1/3`
`b=-1`