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!

$\displaystyle\lim_{n\to \infty} \frac{n(23 + \frac{2}{n})}{n(4+\frac{1}{n})} = \cdots$
Use the fact that $\frac{\alpha}{n}$ tends to $0$ when $n$ tends to infinity, and theorems of limits of sequences. 
