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If lim(xto0) ((4x-1)^(1/3)+a+bx)/(x)=1/3...

If `lim_(xto0) ((4x-1)^(1/3)+a+bx)/(x)=1/3` then find the values of a and b.

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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`
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