Limit of a sequence using $\epsilon - n$ definition I've a problem - I was asked to prove this statement using $\epsilon - n$ definition. 
Here it is : $\displaystyle{\lim_{n \to \infty} \dfrac{n^2+3n}{10-n^2}=-1}$.
My Effort:
Let we have $\epsilon>0$ be given which is fixed but arbitrary. We want to find a $n_0 \in N$, such that $n \ge n_0 \Longrightarrow \left| \dfrac{n^2+3n}{10-n^2} +1 \right| < \epsilon$.
I Noticed that 
$\begin{aligned} \left| \dfrac{n^2+3n}{10-n} +1 \right|&=\left| \dfrac{n^2+3n+10-n^2}{10-n^2} \right| \\ \\ &= \left| \dfrac{3n+10}{10-n^2} \right| \end{aligned}$

Now, I don't know how to proceed. I know I want to make numerator as free of $n$, and denominator as linear expression in $n$ so that I can evaluate $n_0$ by Archimedean Property.
In my book, answer is $n_0 > \max \{ 10, \dfrac{4}{\epsilon} +1 \}$.
I just can't understand why $\max$ came from? 
 A: An easy way of looking at it:
$$\dfrac{n^2+3n}{10-n^2} + 1 = \frac{\frac{10}{n^2}+\frac{3}{n}}{\frac{10}{n^2} - 1}$$
Now, pick a small enough $1 > \epsilon > 0$. Then, by the fact that $\frac{1}{n}$ can be made smaller than any number:

there exists $L$ such that $\frac{10}{l^2} < \epsilon/2$ for all $l > L$.
there exists $J$ such that $\frac{3}{j} < \epsilon/2$ for all $j > J$.
there exists $K$ such that $\frac{10}{k^2} < \epsilon$ for all $k > K$.

Let $N=\max(L,J,K)$.  Then, can you conclude?
A: For large $n$ you have
$$
3n+10<4n
$$
and
$$
n^2-10>\frac{n^2}{2}
$$
Here $n>10$ works. And if you combine these you have
\begin{align*}
\Big |\frac{3n+10}{10-n^2} \Big| &< \Big |\frac{4n}{n^2/2} \Big| \\
&< \Big |\frac{8}{n} \Big| \quad < \quad \epsilon
\end{align*}
If you take $n > \max\{ \frac{8}{\epsilon}, 10\}$ it should work.
Edit: In order for the last inequality to hold you need to have $n>8/\epsilon$. In order for the first simplifier inequalities to hold you need to have $n>10$. We are taking maximum to find the intersection of these requirements.
A: For n>10 $$\lvert\frac{3n+10}{10-n^2}\rvert<\lvert\frac{3n+n}{n-n^2}\rvert=|\frac{4}{1-n}|=ϵ$$
so n=4/ϵ+1, but only for small enough ϵ, otherwise you can take 10 for n
