Proving supremum of set $\left\{\frac{6n-5}{27n - 9n^2 - 20} \middle| n \in \mathbb{N}\right\}$

Question

Prove that $\sup\left\{\frac{6n-5}{27n - 9n^2 - 20} \middle| n \in \mathbb{N}\right\} = 0.$

To prove that supremum of this set is indeed zero, I have to do the following:

1. Prove that $(\forall n \in \mathbb{N})\left(\frac{6n-5}{27n - 9n^2 - 20} \leq 0\right).$ Mine result is that this is true.
2. Prove that $(\forall \varepsilon \in \mathbb{R}^{+})(\exists n \in \mathbb{N})\left(\frac{6n-5}{27n - 9n^2 - 20} > -\varepsilon\right).$

I've got a problem with second statement. How shall I proceed?

Answer 1

Write it as $$\frac{ n\left( 6-\frac{5}{n} \right) }{ -n^2\left( 9-\frac{27}{n}+\frac{20}{n^2}\right) }=-\frac{1}{n} \frac{6-\frac{5}{n}}{ 9-\frac{27}{n}+\frac{20}{n^2}}$$ and show that the second factor can be made arbitrarily close to $\frac{6}{9}$ when $n$ is big. In fact you just need to show that it is bounded, and then the first factor makes the product very close to zero when $n$ is large.