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 $0$?
Thanks in advance.
Best Answer
Hint: Before working with $\varepsilon$, find a number $N\in \mathbf N$ such that $$ \left| \frac{5n^2 -6 }{2n^3 - 7n} \right| \leq \frac{5n^2}{n^3} = \frac{5}{n} \qquad \text{for every} \quad n\geq N.$$