Prove that $\lim\limits_{n \rightarrow \infty} \left(\frac{23n+2}{4n+1}\right) = \frac{23}{4} $.

$\displaystyle\lim_{n\to \infty} \frac{n(23 + \frac{2}{n})}{n(4+\frac{1}{n})} = \cdots$

Use the fact that $\frac{\alpha}{n}$ tends to $0$ when $n$ tends to infinity, and theorems of limits of sequences.

Your proof is basically correct, but I would encourage you to practice a bit on articulating exactly what you mean. Where you say

It is sufficient to show that $$\left| \frac{23n+2}{4n+1} - \frac{23}{4} \right| < \epsilon $$

you mean to say something like

It is sufficient to show that for all $\epsilon\gt0$, there is a $K$ such that for all $n\gt K$, $$\left| \frac{23n+2}{4n+1} - \frac{23}{4} \right| < \epsilon $$

As is, the $\epsilon$ comes out of nowhere and there's no stated restriction on $n$, so the inequality that is "sufficient" to show could be trivially true or patently false.