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
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
Hint:
$ \sqrt[n] {n}\leq \sqrt[n] {9n^2 + 30n + 17}\leq \sqrt[n] {56n^2}. $
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 $\left(\sqrt[n] {n} \right)^3 \to 1$.