Compute the following limit using L'Hospital's rule if appropriate. Use INF to denote infinity and MINF to denote negative infinity.       lim as x approaches infinity of ((((x^3)-(6x^2))^(1/3))-x)

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You need to evaluate the limit, hence you should substitute oo for x in equation under limit such that:

`lim_(x-gtoo) root(3)(x^3 - 6x^2) - x`

You need to multiply and divide by conjugate to `root(3)(x^3 - 6x^2) - x`  such that

`lim_(x-gtoo) (x^3 - 6x^2 - x^3)/(root(3)((x^3 - 6x^2)^2) + x*root(3)(x^3 - 6x^2) + x^2)`

You need to force factor `x^2 ` to denominator such that:

`lim_(x-gtoo) (- 6x^2)/(x^2(root(3)((1 - 6/x)^2) + root(3)(1 - 6/x) + 1)) `

`lim_(x-gtoo) (- 6)/(root(3)((1 - 6/x)^2) + root(3)(1 - 6/x) + 1)`

Substituting oo for x in limit yields:

`lim_(x-gtoo) (- 6)/(root(3)((1 - 6/x)^2) + root(3)(1 - 6/x) + 1) = -6/(root(3)((1 - 6/oo)^2) + root(3)(1 - 6/oo) + 1)`

`lim_(x-gtoo) (- 6)/(root(3)((1 - 6/x)^2) + root(3)(1 - 6/x) + 1) = -6/(1+1+1) = -6/3=-2`

Hence, evaluating the limit to the function yields `lim_(x-gtoo) root(3)(x^3 - 6x^2) - x = -2.`

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