<s>[INST] 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!
 [/INST] 
Is this proof correct? What are some other ways of proving this?
  Thanks!

Your proof is correct with the caveat that you are a bit more precise about what $\epsilon$ and $K$ mean. Another way to prove this is using l'Hôpital's rule. Let $f(n)=23n+2$, $g(n)=4n+1$, then we can see that
$$
\lim_{n\rightarrow\infty} f(n) = \lim_{n\rightarrow\infty} g(n) = \infty
$$
In this case, the rule applies because you have an "indeterminant form" of $\infty/\infty$. Then the rule is that
$$
\lim_{n\rightarrow\infty}\frac{f(n)}{g(n)} = \lim_{n\rightarrow\infty}\frac{f'(n)}{g'(n)}
$$
All that remains is to evaluate $f'(n)=23$, $g'(n)=4$,
$$
\lim_{n\rightarrow\infty}\frac{f(n)}{g(n)} = \lim_{n\rightarrow\infty}\frac{23}{4} = \frac{23}{4}
$$
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