Evaluating $\lim_{n \to\infty}\sqrt[n]{3^n +n^32^n}$ by comparing it with the limits of two other sequences
You have
$$3=\sqrt[n]{3^n}\le \sqrt[n]{3^n +n^32^n}\le\sqrt[n]{3^n +n^33^n}=3\sqrt[n]{1+n^3}.$$
Since the limit of $\sqrt[n]{1+n^3}$ is $1$, you are done.
Hint:
For large $n:$
$3^n< 3^n+n^32^n <2 \cdot 3^n. $
Appended:
$(3/2)^n=$
$e^{n\log (3/2)}>(n^4\log^4 (3/2))/4!$;
Archimedean principle:
There is a $n_0 \in \mathbb{Z^+}$ s.t.
$n_0 >4!/(\log^4 (3/2))$;
For $n\ge n_0$ the inequality
$n^3<(n^4 \log^4 (3/2))/4!<(3/2)^n$ holds.
Used:
$e^x = 1+x+x^2/2!+x^3/3!+.....$