While doing tasks I came across such inequality, near the end of the proof ;
$$n>\frac{\varepsilon(-9\varepsilon + 31)}{27 + 6 \varepsilon}$$
Can we discard some of those epsilon terms or something else?
The problem (former inequality was a result of algebra mistake)is $$\lim_{n \to \infty} \left(\frac{2n-3}{3n+1}\right)^2$$ Proof:
$$\left|\left(\frac{2n-3}{3n+1}\right)^2 - \frac{4}{9}\right| < \varepsilon$$
$$\left|\frac{4n^2 - 12n +9}{9n^2 + 6n + 1} - \frac{4}{9}\right|<\varepsilon$$ $$\left|\frac{4n^2 - 12n + 9 - 36n^2 -24n-4}{81n^2+54n+9}\right|<\varepsilon$$ $$\left|\frac{-32n^2 - 36n + 5}{81n^2 + 54n +9}\right|<\varepsilon$$ Since this expression is always negative, because of absolute value I can multiply it times $(-1)$ and leave absolute value $$\frac{32n^2 + 36n - 5}{81n^2 + 54n + 9}<\varepsilon$$ Here I'm estimating inequality by taking $-5$ from the top of the fraction and $+9$ from denominator That leaves us with $$\frac{32n^2 + 36n}{81n^2 + 54n}<\varepsilon$$ $$\frac{32n + 36}{81n + 54}<\varepsilon$$ $$\frac{32 + \frac{36}{n}}{81 + \frac{54}{n}} < \frac{32 + \frac{36}{n}}{81 } => n > \frac{1}{\varepsilon - \frac{8}{9}}$$ We can go further and discard $\frac{32}{81}$ in the last inequality? Can someone verify the proof?
Your epsilon term on the bottom only makes the bound smaller, so you can get rid of it at the cost of having a slightly larger value of $n$. Then one has $$n> \frac{-9\epsilon^2 +31\epsilon}{27} $$ which isn't so bad. That said, I'm curious what the original problem in question is, because this seems like an unlikely answer for this sort of thing, because as $\epsilon$ gets really small, this quantity goes to 0, where one should expect as $\epsilon$ gets small that the minimum required $n$ should be large. I suspect an algebra mistake somewhere.