Calculate the limit of the given function: lim;x->infinity ((x^3-4x^2-7x-5)^1/3) / ((sqrt x^2-7x)+9) I know that answer is 1, but I don't know how to solve it. Thank's!

Expert Answers

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You should force factor `x^3`  to numerator and `x^2 ` to denominator, such that:

`lim_(x-gtoo)(root(3)(x^3(1 - (4x^2)/(x^3) - (7x)/(x^3) - 5/(x^3))))/(sqrt(x^2(1 - (7x)/(x^2)))+9)` = `lim_(x->oo)(xroot(3)(1 - 4/x - 7/x^2 - 5/x^3))/(x(sqrt(1 - 7/x)+9/x))`

Reducing like terms yields:

`lim_(x->oo)(root(3)(1 - 4/x - 7/x^2 - 5/x^3))/(sqrt(1 - 7/x)) = (root(3)(1 - lim_(x->oo)4/x - lim_(x->oo)7/x^2 - lim_(x->oo)5/x^3))/(sqrt(1 - lim_(x->oo)(7/x) + lim_(x->oo)9/x)`

`lim_(x->oo)(root(3)(1 - 4/x - 7/x^2 - 5/x^3))/(sqrt(1 - 7/x)) = (root(3)(1 - 0 - 0 - 0))/(sqrt(1 - 0)+0) `

`lim_(x->oo)(root(3)(1 - 4/x - 7/x^2 - 5/x^3))/(sqrt(1 - 7/x)) = 1/1`

`lim_(x->oo)(root(3)(1 - 4/x - 7/x^2 - 5/x^3))/(sqrt(1 - 7/x)) = 1`

Hence, evaluating the limit yields `lim_(x->oo)(root(3)(x^3-4x^2-7x-5))/(sqrt(x^2-7x)+9)= 1.`

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