Absolute value in sequence proof

Question

That's very basic question, but I want to know how to approach such problem because it's unclear to me - In sequence limit proof (Epsilon N) in case like this $$\left|\frac{6}{3-n}\right|<\varepsilon$$ term on the left isn't always positive so we have to look at expanded inequality, but then (I checked) we get these two inequalities: $$n < 3+\frac{6}{\varepsilon} $$ $$n<3-\frac{6}{\varepsilon}$$ Can someone explain to me what do we do with $N_\varepsilon$ and maybe I'm wrong with operations?

Answer 1

As you look for the limit, you should assume that your integer $ n$ is great. so, $$|3-n|=-(3-n)=n-3$$

and the condition $$|\frac{6}{3-n}|<\epsilon$$ becomes $$\frac{6}{n-3}<\epsilon$$ or $$n>3+\frac{6}{\epsilon}$$

So, you can take $$N = \lfloor \frac{6}{\epsilon}\rfloor +4$$