<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] Personally (assuming I wasn't supposed to use sharper tools), I would tackle it by first noting that the fraction should be about $1/n$ when $n$ is very large. So pull out a factor $1/n$, and simplify the rest, and you get
$$\frac{n+6}{n^2-6}=\frac1n\cdot\frac{1+6/n}{1-6/n^2}.$$
Now you just need to find a good bound for the fracton on the right, when $n$ is large. To make life really simple, you could require $n\ge6$, which implies $1+6/n\le2$ and $1-6/n^2\ge5/6$, so that
$$\frac{n+6}{n^2-6}\le\frac{12}5\frac1n\qquad\text{for $n\ge6$}.$$
The rest is now trivial.
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