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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.

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

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Use $5n^{2} -6 <5n^{2}$ and $2n^{3}-7n > n^{3}$ if $n >3$.

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