This is more of a solution verification request than anything else. Feedback would be much appreciated.
Question: Prove that the sequence $\{a_n\}_{n=1}^\infty$ defined by $a_n=\frac{n+37}{n^5-2n^2}$ converges to $0$ using the $\varepsilon$-$N$ definition of a limit.
Solution: Need to show that $\forall\epsilon>0$, $\exists N\in\mathbb{N}$ such that $n>N \implies|a_n-0|<\varepsilon.$
Fix $\epsilon>0$.
We have $\left|a_n-0\right|=\left|\frac{n+37}{n^5-2n^2}\right|$.
Now $n+37\le n+37n=38n$ provided $n \ge1$.
Likewise $n^5-2n^2\ge n^5-\frac{1}{2}n^5=\frac{1}{2}n^5$ provided $\frac{1}{2}n^5\ge 2n^2$, which is true provided $n^3\ge 4$ i.e when $n\ge 2$.
Then we have $\left|\frac{n+37}{n^5-2n^2}\right|\le\left|\frac{38n}{\frac{1}{2}n^5}\right|=\left|\frac{76}{n^4}\right|=\frac{76}{n^4}$
Then $\frac{76}{n^4}<\epsilon \iff\frac{n^4}{76}>\frac{1}{\epsilon}\iff n>\sqrt[4]\frac{76}{\epsilon}$
So choose $N=\lfloor\sqrt[4]\frac{76}{\epsilon}\rfloor+1$ and the definition is satisfied. Q.E.D.