dream13rxs

2022-07-02

How do I find
$\underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{{3}^{n}+{n}^{3}{2}^{n}}$
by comparing it with other two sequences?

toriannucz

Expert

You have
$3=\sqrt[n]{{3}^{n}}\le \sqrt[n]{{3}^{n}+{n}^{3}{2}^{n}}\le \sqrt[n]{{3}^{n}+{n}^{3}{3}^{n}}=3\sqrt[n]{1+{n}^{3}}.$
Since the limit of $\sqrt[n]{1+{n}^{3}}$ is 1, you are done.

Do you have a similar question?