How do I prove that limit by definition?

Question

How do I prove that $\lim\limits_{n\to \infty}\left(\frac{6n^2-8}{n^2-8}\right)=6$, by definition? I need to find $N_{\epsilon}$ such that for every $n>N_{\epsilon}$: $$\left|\left(\frac{6n^2-8}{n^2-8}\right)-6\right|<\epsilon.$$ How do I find the $N_{\epsilon}$?

Answer 1

For this sort of question, it is usually helpful to pretend that you have already had the desired inequality, and you solve the threshold out of the inequality.

Namely, you want to find $N_\varepsilon$ such that $$ \left|\left(\frac{6n^2-8}{n^2-8}\right)-6\right|<\varepsilon, $$ whenever $n>N_\varepsilon$.

Now, just try to separate $n$ and let us assume $n>2$: \begin{align} \left|\left(\frac{6n^2-8}{n^2-8}\right)-6\right|<\varepsilon &\iff \left|\frac{40}{n^2-8}\right|<\varepsilon\\ &\iff n^2-8>\frac{40}{\varepsilon}\\ &\iff n^2>8+\frac{40}{\varepsilon}. \end{align} Thus, you can just take $$ N_\varepsilon=\left\lfloor\sqrt{8+\frac{40}{\varepsilon}}\right\rfloor+1 $$ and we are done.