<s>[INST] Proof of limit using epsilon-L proof We have $\lim \frac{n+6}{n^2-6}=0$. I need to prove this limit using the epsilon-L proof method.
I wrote down the following proof:
$\dfrac{n+6}{n^2-6}<\dfrac{2n}{\frac{1}{2}n^2}=\dfrac{4}{n}<\epsilon$ (why are we allowed to say this?) so $n>\dfrac{4}{\epsilon}$.
Proof. Let $\epsilon>0$ and let $N=\max\{4,\dfrac{4}{\epsilon}\}$ (since the inequality is only valid for $n>6$ right? Since $n+6<2n$ gives $n>6$ and $n^2-6>\frac{1}{2}n^2$ gives $n>4$)?? I don't understand this too.
Can anyone clarify? I can't seem to understand anything about this *** proof
 [/INST] For $n>6$, we have
$$n^2-6 >n^2-36>0$$
$$n^2-36=(n+6)(n-6) $$
thus
$$\frac {n+6}{n^2-6}<\frac {1}{n-6} $$
To satisfy $\frac {n+6}{n^2-6}<\epsilon $, it sufficient to have
$n>6$ and $\frac {1}{n-6}<\epsilon $
or $n>6$ and $n>6+\frac {1}{\epsilon} $
so we will take $N=6+\frac {1}{\epsilon} $.
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