# Proof that the sequence $\left\{\frac{5n^2-6}{2n^3-7n}\right\}$ converges to $0$

We're asked to prove that the sequence $$\left\{\frac{5n^2-6}{2n^3-7n}\right\}$$ converges to $$0$$.

Attempt:

We need to show that $$\left|\frac{5n^2-6}{2n^3-7n}-0\right|<\epsilon\rightarrow\left|\frac{5n^2-6}{2n^3-7n}\right|<\epsilon$$ such that we need to find an upper bound for the numerator and a lower bound for the denominator. Thus $$\forall n>2$$ we have $$5n^2-6<2n^3\\ 2n^3-7n>\frac{1}{4}n^2\\ \frac{2n^3}{\frac{1}{4}n^2}=8n<\epsilon$$

I feel like am lost half-way through my own proof, isn't there a way to suppose that this limit is less than another limit converging into $$0$$ thus implying it also must converge to $$0$$?

• Lim for $n \to$ what? Oct 22 '18 at 8:54
• $n\rightarrow{}\infty$ Forgot to mention that Oct 22 '18 at 9:14
Hint: Before working with $$\varepsilon$$, find a number $$N\in \mathbf N$$ such that $$\left| \frac{5n^2 -6 }{2n^3 - 7n} \right| \leq \frac{5n^2}{n^3} = \frac{5}{n} \qquad \text{for every} \quad n\geq N.$$
Use $$5n^{2} -6 <5n^{2}$$ and $$2n^{3}-7n > n^{3}$$ if $$n >3$$.