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!


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 $f'(n)=23$, $g'(n)=4$,
$$
\lim_{n\rightarrow\infty}\frac{f(n)}{g(n)} = \lim_{n\rightarrow\infty}\frac{23}{4} = \frac{23}{4}
$$
