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How can I use the fact that $\sqrt[n] {n} \to 1 $ in order to calculate the limit of the sequence: $ \sqrt[n] {9n^2 + 30n + 17} $ ?

Thanks

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2 Answers 2

up vote 1 down vote accepted

Hint:

$$ \sqrt[n] {n}\leq \sqrt[n] {9n^2 + 30n + 17}\leq \sqrt[n] {56n^2}. $$

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can you please explain how you got these bounds? –  yuta Nov 23 '12 at 16:12
    
One of them is easy. For the other one I used that $n\leq n^2$ and $1\leq n^2$. Thus $rn\leq rn^2$ and $r\leq rn^2 \ \ \forall r,n \in \mathbb{N}$ . –  P.. Nov 23 '12 at 16:16
    
Wow ! So Easy! Thanks a lot ! –  yuta Nov 23 '12 at 16:36

For $n \gt 0$ clearly $\sqrt[n] {9n^2 + 30n + 17} \gt 1$.

Meanwhile for $n \ge 31$ you have $30n + 17 \lt n^2$ so $$\sqrt[n] {9n^2 + 30n + 17} \lt \sqrt[n] {10 n^2 } \lt \sqrt[n] {n^3 } = \left(\sqrt[n] {n} \right)^3$$ and since $\sqrt[n] {n} \to 1$ you also have $\left(\sqrt[n] {n} \right)^3 \to 1$.

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