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I've been trying to show this unfortunate polynomial converges through epsilon proof, without using any convenient theorems. I think I have found a simple way to handle it, however, I am not sure about the last steps, specifically showing $\frac{2}{n^2}$ is less than $\epsilon$.

Show that $\frac{2n^2 + 2n}{n^3 + 4n^2 + 11} \rightarrow 2, n \rightarrow \infty$

Given $\epsilon > 0, \exists N \ s.t. \ \forall n \geq N$,

$|\frac{2n^2 + 2n}{n^3 + 4n^2 + 11} - 2| = |\frac{2n - 8n^2 - 22}{n^3 + 4n^2 + 11}| < |\frac{2n - 8n^2 - 22}{n^3}| < |\frac{2n}{n^3}| = \frac{2}{n^2}$

Let $N = \frac{1}{\sqrt{\epsilon}}$, since $n \geq N$ we know $\frac{1}{n} < \frac{1}{N} = \sqrt{\epsilon}$, so we have

$\frac{2}{n} * \frac{1}{n} <2 * \frac{1}{n} * \sqrt{\epsilon} < 2\sqrt{\epsilon}\sqrt{\epsilon} < 2\epsilon $

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I would do that last part in reverse to show how the bound on n is derived. This also reduces the chance of error. As you can see, I came up with a slightly different bound.

$2/n^2 < c \iff n^2 > 2/c \iff n> \sqrt{2/c} $.

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    Sorry, I am not sure I follow. If you choose $c = 2\epsilon$ wouldn't you get the same result as me $\frac{1}{n} < \sqrt{\frac{c}{2}}$? – 2017-01-27
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    I use $c$ instead of $\epsilon$ since it is easier to enter. Also, I always start with the expression (in this case, $2/N^2$) that is a bound on the error and, from that, derive the bound on $N$. I do not like magically coming up with the bound on $N$ and then verifying that it works. – 2017-01-28
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    Ah I see, so in my case what I did was correct, but in general it is better to derive the bound rather than choose it like I did. Thank you! – 2017-01-28