Calculate the limits of the sequence $\frac{2}{n+2}$ from first principles

Question

i know that it can then be simplified by saying $\frac{2}{n+2}$ < $\frac{2}{n}$ but then would it just continue as normal saying that you would then choose an integer $N(\epsilon)$ within the set of natural numbers such that $$N(\epsilon) > \frac{2}{\epsilon} $$ possible by Archimedes principle. if someone could talk me through this it would be much appreciated.

Answer 1

We have $\lim_{n\to\infty}\frac{2}{n+2}=0$. Here, the inequality $|\frac{2}{n+2}-0|<\varepsilon$ gives $n > \frac{2-2\varepsilon}{\varepsilon}$ and hence any $N >\frac{2(1-\varepsilon)}{\varepsilon}$ will do.