$\epsilon-N$ Proof of sequence convergence $\{\frac{3+5n^2}{n+n^2}\mapsto 5\}$ Here's what I have so far:
$|\frac{3+5n^2}{n+n^2}-5| &lt; |\frac{3}{n}-5|$.
I don't know what to do from here.
I can't see a good use for the inequality you produced. I would evaluate:
$$\bigg|\frac{3+5n^2-5(n+n^2)}{n+n^2}\bigg|=\bigg|\frac{3-5n}{n+n^2}\bigg|=\bigg|\frac{\frac3n-5}{1+n}\bigg|$$
As $n\to \infty, \frac 3n$ vanishes, so we see that $$\bigg|\frac{\frac3n-5}{1+n}\bigg|&lt;\bigg|\frac{-5}{1+n}\bigg|&lt;\bigg|\frac{-5}{n}\bigg|$$
To see this, notice that $n&gt;0\implies \frac 3n-5&lt;0\implies |\frac3n-5|=5-\frac 3n$, and so we have added to the numerator, thus getting a larger number in the first step. In the second step, we subtract from the denominator (i.e. divide by a smaller number), so we again get a larger number.
Then, we have $\bigg|\frac 5n\bigg|=\frac5n&lt;\epsilon\implies n&gt;\frac 5\epsilon$, so take $N=\lceil\frac 5\epsilon\rceil$
$\bigg|\frac{3+5n^2}{n+n^2}-5\bigg| = \bigg|\frac{3/n^2+5}{(1/n+1)}-\frac{5(1/n+1)}{{(1/n}+1)}\bigg|$ = $\bigg|\frac{3/n^{2}-5/n}{1/n+1}\bigg| = \bigg|\frac{3/n-5}{1+n}\bigg| = \bigg|{\frac{(3/n)}{(1+n)} - \frac{5}{1+n}}\bigg| \leq \bigg|{\frac{(3/n)}{(1+n)}\bigg| + \bigg|\frac{5}{1+n}}\bigg|$ by a corollary of the triangle inequality. To make both $\bigg|{\frac{(3/n)}{(1+n)}}\bigg|$ and $\bigg|{\frac{5}{1+n}}\bigg| &lt; \frac{\epsilon}{2}$. We need ${\frac{5}{1+n}} &lt; \frac{\epsilon}{2}$ and the necessary $n$ is found by simple algebra. For $ {\frac{(3/n)}{(1+n)}} &lt;\frac{\epsilon}{2}$ we get $\frac{6}{\epsilon} &lt; n^{2} + n^{3}$ so it is enough to make $\frac{6}{\epsilon} &lt; n^{2} + n^{2}=2n^{2}$. Then we get the necessary $n$.  Since both of these are requirements to satisfy $\bigg|{\frac{(3/n)}{(1+n)}}\bigg| + \bigg|{\frac{5}{1+n}}\bigg| &lt; \epsilon,$ choose $n &gt; \max\{\frac{10-\epsilon}{\epsilon},\sqrt\frac{3}{\epsilon}\} $.