<s>[INST] Prove that $ \lim_{n \rightarrow \infty } \frac{n+6}{n^2-6} = 0 $. My attempt:
We prove that $ \lim\limits_{n \rightarrow \infty }  \dfrac{n+6}{n^2-6} =0$. 
It is sufficient to show that for any $ \epsilon \in\textbf{R}^+ $, there exists an $ K \in \textbf{R}$ such that for any $ n > K $  
$ \displaystyle \left| \frac{n+6}{n^2-6} -  0 \right| <  \epsilon $.  
Suppose $ n>4  $.  Then $ n+6 < 7n $ and $ n^2 -6 > \frac{1}{2} n^2 $.  So 
$ \displaystyle \left| \frac{n+6}{n^2-6} -  0 \right| = \frac{n+6}{n^2-6 } < \frac{7n}{\frac{1}{2}n^2} = \frac{14}{n} $. 
Consider $ K = \max\{4,\displaystyle \frac{14}{\epsilon}\} $ and suppose $ n> K $.  Then $ n > \displaystyle \frac{14}{\epsilon} $.  This implies that $ \epsilon > \displaystyle \frac{14}{n} $.  Therefore     
$ \displaystyle \left| \frac{n+6}{n^2-6} -  0 \right| = \frac{n+6}{n^2-6 } < \frac{7n}{\frac{1}{2}n^2} = \frac{14}{n} < \epsilon $.  
Thus  $ \lim\limits_{n \rightarrow \infty } \displaystyle \frac{n+6}{n^2-6}=0 $. 
Is this proof correct? What are some other ways of proving this? Thanks!
 [/INST] Your proof is correct. Just to be pedantic, you can define $K$ directly as $14/\varepsilon$, since you precedently assumed that $n>4$. 
As you've already noted in the comments, the actual way to solve limits without going through all those $\delta$-$\epsilon$ arguments is to use limits properties, such as the preservation of operations $+,\cdot$ (and, with some attention, $\div$). With these little theorems the proof is very easy: $$\{\dfrac{n+6}{n^2-6}\}_{n\to \infty}=\{\dfrac{1+6/n}{n-6/n}\}_{n\to \infty}=\dfrac{ \lim 1 + \lim 6/n}{\lim n -\lim6/n}=0$$
since all but the left bottom corner terms are bounded.
</s>