Prove that $\lim\limits_{n \rightarrow \infty} \left(\frac{23n+2}{4n+1}\right) = \frac{23}{4} $. My attempt:   
We prove that $$\displaystyle \lim\limits_{n \rightarrow \infty} \left(\frac{23n+2}{4n+1}\right) = \frac{23}{4} $$ 
It is sufficient to show that for an arbitrary  real number  $\epsilon\gt0$, there is a $K$
 such that for all $n\gt K$, $$\left| \frac{23n+2}{4n+1} - \frac{23}{4}
 \right| < \epsilon.  $$
Note that $$ \displaystyle\left| \frac{23n+2}{4n+1} - \frac{23}{4} \right| = \left| \frac{-15}{16n+4} \right|  $$ and for $ n > 1 $ $$ \displaystyle \left| \frac{-15}{16n+4} \right| = \frac{15}{16n+4} < \frac{1}{n}. $$  
Suppose $ \epsilon \in \textbf{R} $ and $ \epsilon > 0  $.  Consider $ K = \displaystyle \frac{1}{\epsilon} $.  Allow that $ n > K $.  Then $ n > \displaystyle \frac{1}{\epsilon} $.  So $ \epsilon >\displaystyle \frac{1}{n} $.
Thus 
$$ \displaystyle\left| \frac{23n+2}{4n+1} - \frac{23}{4} \right| = \left| \frac{-15}{16n+4} \right| = \frac{15}{16n+4} < \frac{1}{n} < \epsilon. $$  Thus $$ \displaystyle \lim\limits_{n \rightarrow \infty} \left(\frac{23n+2}{4n+1}\right) = \frac{23}{4}. $$  
Is this proof correct? What are some other ways of proving this? Thanks!

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.
