<s>[INST] Epsilon-N method - Proof verification 
Prove (using the epsilon-N method) that the sequence of numbers $\dfrac{5n^3-2}{n^3}$ converges. Calculate the limit first. 

First we calculate the limit: $\lim_{n \to \infty} \dfrac{5n^3-2}{n^3} = \lim_{n \to \infty} 5- \dfrac{2}{n^3}  = 5$. So we are required to show, for a given $\epsilon > 0 $, that there exists a positive integer $N$ such that if $n>N$, then 
$|(5-\dfrac{2}{n^3}) -5 | = |\dfrac{-2}{n^3} | = \dfrac{2}{n^3} < \epsilon $
$$ \dfrac{2}{n^3} < \epsilon \to n > \sqrt[3]{\dfrac{2}{\epsilon}}$$
Therefore, if we choose $N = \lceil \sqrt[3]{\dfrac{2}{\epsilon}} \rceil $, and we let $n$ be any integer such that $n>N$, then $n> \sqrt[3]{\dfrac{2}{\epsilon}}$.
Now we can set up the proof:
Let $\epsilon > 0$. Choose $N = \lceil \sqrt[3]{\dfrac{2}{\epsilon}} \rceil $ and let $n$ be any integer such that $n>N$. Thus $n > \sqrt[3]{\dfrac{2}{\epsilon}}$ and $\epsilon > \dfrac{2}{n^3}$. Therefore
$$|(5-\dfrac{2}{n^3}) -5 | = |\dfrac{-2}{n^3} | = \dfrac{2}{n^3} < \epsilon $$
Thus we conclude that the sequence of numbers $\dfrac{5n^3-2}{n^3}$ converges. 
 [/INST] The proof looks correct. The only thing I noticed was your use of the ceiling function $\lceil$ and $\rceil$. 
Unless your professor taught limits in this manner, traditionally one might invoke the Archimedean Principle to show that there exists $N \in \mathbb{N}$ such that $N > a$ for some real number $a$. In fact, the proof of the ceiling function is usually shown as a basic corollary to this principle.
The use of the ceiling function is certainly not incorrect, just make sure that is okay to use in your proof.
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