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!