Use an $\epsilon-N$ argument to find and prove the $\lim_{n\rightarrow\infty}\sqrt[n]{5+n^2}$. Try some variations of your own.
I have to find and prove this limit using $\epsilon-N$ argument.
I think that the limit is 1, since the limit as n tends to infinity of the nth root of a polynomial(in this case it is $5+n^2$) is 1
Let's fix some $\varepsilon > 0$, and consider when $|\sqrt[n]{5 + n^2} - 1| < \varepsilon$. Note that $\sqrt[n]{5 + n^2} - 1 > 0$, so this is equivalent to finding where $$\sqrt[n]{5 + n^2} < 1 + \varepsilon \iff 5 + n^2 < (1 + \varepsilon)^n.$$ Both sides should tend to $\infty$, though the right tends eventually much faster. We can see this using the binomial theorem (for $n \ge 2$): \begin{align*} (1 + \varepsilon)^n &= 1 + n\varepsilon + \frac{1}{2}n(n+1)\varepsilon^2 + \ldots \\ &\ge 1 + n\varepsilon + \frac{1}{2}n(n+1)\varepsilon^2 \\ &\ge 1 + n\varepsilon + \frac{1}{2}n^2\varepsilon^2. \end{align*} So, if we can find $N$ such that $$n > N \implies 5 + n^2 < 1 + n\varepsilon + \frac{1}{2}n^2\varepsilon^2,$$ then we are done.
In fact, choose an $N \ge \frac{\sqrt{2}}{\sqrt{\varepsilon}}$, and you'll get that $\frac{1}{2}n^2 \varepsilon^2 > n^2$. If you choose $N \ge \frac{4}{\varepsilon}$, then $1 + n \varepsilon > 5$. Plus, of course, we must have $n \ge 2$ for our binomial theorem analysis to work. So, our choice of $N$ could be $$N = \max\left\{2, \frac{4}{\varepsilon}, \frac{\sqrt{2}}{\sqrt{\varepsilon}}\right\}.$$
Of course, now you should assemble this into a direct proof!