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I'm trying to show that $\exists \,\varepsilon >0\mid\forall n>N\in\mathbb{N}$ such that: $$\left|\frac{6n^3+5n-1}{2n^3+2n+8}-3\right| < \varepsilon$$ Let's take $\varepsilon = 1/2$:

$$\left|\frac{6n^3+5n-1}{2n^3+2n+8}-3\right| = \left|\frac{6n^3+5n-1 - 6n^3 -6n -24}{2n^3+2n+8}\right| < \left|\frac{-n -25}{2n^3+2n+8}\right| = \left|-\left(\frac{n +25}{2n^3+2n+8}\right)\right|$$

Since $|-x| = |x|$:

$$\left|-\left(\frac{n +25}{2n^3+2n+8}\right)\right| = \left|\frac{n +25}{2n^3+2n+8}\right|<\varepsilon $$

This is the part which I have trouble with. Here, I always end up with my minimum-required index to be smaller than zero, which seems peculiar to me:

$$\left|\frac{n +25}{2n^3+2n+8}\right| < |n+25| = n+ 25 < \varepsilon = 1/2$$ From here, I will get that my $N$ will be less than zero, which means that all the elements of the sequence are inside of my given epsilon environment, but I know that's not true since $a_1 = 5/6 < 3 - \varepsilon $, so what I did do wrong? Is it because I completely removed the denominator? If so, why does that break the inequality?

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    $\begingroup$ Yes $\left|\frac{n+25}{2n^3+2n+8}\right|\lt|n+25|$ but this will not allow you to solve the limit in question. I suggest writing instead $\left|\frac{n+25}{2n^3+2n+8}\right|\lt|\frac{1}{n^2}|$ for large enough $n$. $\endgroup$ – Peter Foreman Apr 16 at 21:54
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    $\begingroup$ You don't want to show that there exists $\varepsilon$, you want to show that any positive $\varepsilon$ satisfies the property. $\endgroup$ – Git Gud Apr 16 at 21:58
  • $\begingroup$ See this for a similar example, and also for general strategy. $\endgroup$ – Git Gud Apr 16 at 21:59
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Don't worry about $\epsilon$. Instead, try to get the difference in a simple form by assuming $n$ is as large as you need. Then getting $n$ is much simpler.

Using your calculations:

$\begin{array}\\ \left|\dfrac{6n^3+5n-1}{2n^3+2n+8}-3\right| &= \left|\dfrac{6n^3+5n-1 - 6n^3 -6n -24}{2n^3+2n+8}\right|\\ &= \left|\dfrac{-n -25}{2n^3+2n+8}\right|\\ &= \left|\dfrac{n +25}{2n^3+2n+8}\right|\\ &\le \left|\dfrac{2n}{2n^3}\right| \qquad\text{if } n \ge 25\\ &= \left|\dfrac{1}{n^2}\right|\\ &\lt \epsilon \qquad\text{if } n \gt \frac1{\sqrt{\epsilon}}\\ \end{array} $

Therefore the difference is within $\epsilon$ if $n \gt \max(25, \frac1{\sqrt{\epsilon}})$.

You don't have to get the best possible $n$ - just showing one exists is good enough.

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  • $\begingroup$ Well, I worry a little bit about $\varepsilon$, when the first sentence of the OP starts with "$\exists \,\varepsilon >0\mid\forall n>N\in\mathbb{N}$". $\endgroup$ – Jean-Claude Arbaut Apr 16 at 22:12
  • $\begingroup$ $N = 1/\sqrt{\epsilon}$. $\endgroup$ – marty cohen Apr 17 at 4:16
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Note that for $n\geq 1$ we have $n+25\leq n+25n=26n$ and $2n^3+2n+8\geq 2n^3$ so you have that $$\left|\frac{n+25}{2n^3+2n+8}\right|\leq \frac{26n}{2n^3}=\frac{13}{n^2}$$

Can you take it from here?

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I don't understand why everyone is making this so long and drawn out.

You only need remember this: as $n\to \infty$; $t\geq0\implies n^{t+1}-n^t\to\infty$

In other words, given $\lim_{n\to\infty}(\frac{P(n)}{Q(n)})$ where $P$ and $Q$ are polynomials, only the largest powers (as long as they are positive) and their coefficients count.

So: $$\lim_{n\to\infty}\bigg[\frac{6n^3+5n-1}{2n^3+2n=8}\bigg]=\lim_{n\to\infty}\bigg[\frac{6n^3}{2n^3}\bigg]=\lim_{n\to\infty}[3]=3$$

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I do it like this:

$\forall n \ge 1, \; \dfrac{6n^3 + 5n - 1}{2n^3 + 2n + 8} = \dfrac{6 + 5n^{-2} - n^{-3}}{2 + 2n^{-2} + 8n^{-3}}; \tag 1$

since it is easy to see that

$\displaystyle \lim_{n \to \infty} (6 + 5n^{-2} - n^{-3}) = 6, \tag 2$

and

$\displaystyle \lim_{n \to \infty} (2 + 2n^{-2} + 8n^{-3} ) = 2, \tag 3$

we have

$\displaystyle \lim_{n \to \infty} \dfrac{6n^3 + 5n - 1}{2n^3 + 2n + 8} = \lim_{n \to \infty} \dfrac{6 + 5n^{-2} - n^{-3}}{2 + 2n^{-2} + 8n^{-3}} = \dfrac{\lim_{n \to \infty}( 6 + 5n^{-2} - n^{-3})}{\lim_{n \to \infty} ( 2 + 2n^{-2} + 8n^{-3}) } = \dfrac{6}{2} = 3. \tag 4$

Of course, such an answer, though perfectly rigorous and on point, is bound to be less than satisfactory to readers who want, as the text of the question indicates, to see the $\epsilon$-$N$ mechanism operate in detail; what I have done here is simply invoke standard and elementary results on the behavior of limits; we may also cast things into $\epsilon$-$N$ form if we write

$\dfrac{6 + 5n^{-2} - n^{-3}}{2 + 2n^{-2} + 8n^{-3}} - 3$ $= \dfrac{ 6 + 5n^{-2} - n^{-3} - 3( 2 + 2n^{-2} + 8n^{-3})}{2 + 2n^{-2} + 8n^{-3}}$ $=\dfrac{5n^{-2} - n^{-3} - 6n^{-2} - 24n^{-3}}{2 + 2n^{-2} + 8n^{-3}} = \dfrac{-n^{-2} - 25n^{-3}}{2 + 2n^{-2} + 8n^{-3}} ; \tag 5$

$\left \vert \dfrac{6 + 5n^{-2} - n^{-3}}{2 + 2n^{-2} + 8n^{-3}} - 3 \right \vert = \left \vert \dfrac{-n^{-2} - 25n^{-3}}{2 + 2n^{-2} + 8n^{-3}} \right \vert; \tag 6$

it is evident via inspection of the right-hand side of this equation that, given any $\epsilon > 0$ there exists a sufficiently large $N \in \Bbb N$ that

$n > N \Longrightarrow \left \vert \dfrac{-n^{-2} - 25n^{-3}}{2 + 2n^{-2} + 8n^{-3}} \right \vert < \epsilon, \tag 7$

which in light of (6) by definition implies

$\displaystyle \lim_{n \to \infty} \dfrac{6 + 5n^{-2} - n^{-3}}{2 + 2n^{-2} + 8n^{-3}} = 3, \tag 8$

completing an $\epsilon$-$N$ demonstration of the requisite limit.

It appears to me that our OP Fatima Ens' principal error lies in confusing the roles of $\exists$ and $\forall$ in the definition of limit; the quantifying symbol string should read

$\forall \epsilon \exists N \mid \forall n > N \; \text{and so forth}; \tag 9$

if this correction is accepted and the remaining statements bought into accord with this (corrected) version, the chances of completing a successful proof are greatly enhanced.

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