Prove that $$\lim \frac{n+6}{n^2-6}=0$$ using $\epsilon-\delta$
Given an $\epsilon >0$, we need to find an $n_0 \in \mathbb{N}$ such that $|a_n|<\epsilon$, $\forall$ $n \geq n_0$.
I started with triangle inequality
$$|n^2-6|\leq n^2+6$$
$\implies$
$$\epsilon >\frac{n+6}{n^2-6}\geq \frac{n+6}{n^2+6}$$
$\implies$
$$n^2+6 > \frac{n+6}{\epsilon}$$
$\implies$
$$n^2-\frac{n}{\epsilon}+6-\frac{6}{\epsilon} > 0$$
By Quadratic formula we get
$$n > \frac{1+\sqrt{1+24\epsilon-24\epsilon^2}}{2\epsilon}$$
So we found that $$n_0=\left \lceil \frac{1+\sqrt{1+24 \varepsilon-24 \varepsilon^{2}}}{2 \varepsilon} \right \rceil $$
now we can prove the actual claim backwards. But my doubt is, what if $\epsilon >1$ is chosen?
because when $\epsilon >1$, $1+24\epsilon-24\epsilon^2 <0$
Best Answer
We can find $n_1$ such that for any $n> n_1$, $|a_n|<1$.
Hence if $\epsilon > 1$, we can use the same $n_1$ because if $n > n_1$, then we have $|a_n| < 1 < \epsilon$.