A simple limit of a sequence I feel almost ashamed for putting this up here, but oh well..
I'm attempting to prove: $$\lim_{n\to \infty}\sqrt[n]{2^n+n^5}=2$$
My approach was to use the following inequality (which is quite easy to prove) and the squeeze theorem:
$$\lim_{n\to \infty}\sqrt[n]{1}\leq \lim_{n\to \infty}\sqrt[n]{1+\frac{n^5}{2^n}}\leq \lim_{n\to \infty}\sqrt[n]{1+\frac{1}{n}}$$
I encountered a problem with the last limit though. While with functions using the $e^{lnx}$ "trick" would work, this problem is a sequences problem only, therefore there has to be a solution only using the first-semester-calculus-student-who's-just-done-sequences knowledge.
Or maybe this approach to the limit is overly complicated to begin with? Anyway, I'll be glad to receive any hints and answers, whether to the original problem or with $\lim_{n\to \infty}\sqrt[n]{1+\frac{1}{n}}$, thanks!

The other answers provide easier approachs to your limit, that I would prefer. But the limit for your upper bound can be done straightforwardly: by definition,
$$
\sqrt[n]{1+\frac1n}=\left(1+\frac1n\right)^{1/n}=e^{\frac1n\,\log(1+\frac1n)}.
$$
As the exponential is continuous, all you need to do is to calculate the limit of $\frac1n\,\log(1+\frac1n)$, and this is trivial as it is a product where both factors go to zero.
