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## Homework Statement

For[tex]Q(x) = x^k + \sum_{n=0}^{k-1} a_n x^n[/tex]

Find [tex]\lim_{x→∞}\left({Q(x)^{\frac{1}{k}} - x}\right)[/tex]

## Homework Equations

L'Hopital

## The Attempt at a Solution

[tex]Q(x) - x^k = \sum_{n=0}^{k-1} a_n x^n[/tex]

[tex]Q(x) - x^k = (Q(x)^{\frac{1}{k}})^k - x^k = \left(Q(x)^{\frac{1}{k}} - x\right)\left([Q(x)^{\frac{1}{k}}]^{k-1} + x[Q(x)^{\frac{1}{k}}]^{k-2} + ... + x^{k-2}Q(x)^{\frac{1}{k}} + x^{k-1} \right)[/tex]

[tex]Q(x)^{\frac{1}{k}} - x = \frac{Q(x) - x^k}{[Q(x)^{\frac{1}{k}}]^{k-1} + x[Q(x)^{\frac{1}{k}}]^{k-2} + ... + x^{k-2}Q(x)^{\frac{1}{k}} + x^{k-1}}[/tex]

Then [itex]\lim_{x→∞}{\left(Q(x)^{\frac{1}{k}} - x\right)} = \lim_{x→∞}\frac{f(x)}{g(x)}[/itex], and if I can show [itex]\lim_{x→∞}g(x) = ∞[/itex], I can use L'Hopital.

Am I headed in the right direction? I can see this either working out nicely after a lot of work or being an incredible waste of time. Am I overlooking a simpler way to evaluate this limit?