Proving a limit exists - solving for epsilon with absolute values I have the equation that I want to prove the limit goes to 1:
$$\lim_{n \to \infty} \frac {(n+8)(n+1)}{n(n-10)} = 1$$
Using definition of limit, I get this equation:
$$ \left | \frac {(n+8)(n+1)}{n(n-10)} - 1 \right| \lt \epsilon $$
which I can manipulate into the following using: $ 1 = \frac {n^2 - 10n}{n^2-10n}$:
$$ \left | \frac {n^2 + 9n + 8 - n^2 + 10n}{n^2 - 10n} \right | = \left | \frac{19n + 8}{n^2 - 10n} \right | \lt \epsilon $$
Any suggestions on how to solve this for epsilon? Sorry if this is simple, but I am not sure what concept to search for. 
References or suggestions about what to search are also welcome.
 A: I assume you have to use the $\epsilon$-$N$ definition of the limit instead of other methods, correct? 
If so, you shouldn't be trying to find an $\epsilon$. 
For a given $\epsilon > 0$, you need to find an $N$ such that $\left|\dfrac{(n+8)(n+1)}{n(n-10)}-1\right| < \epsilon$ for all $n > N$. 
So, you want $\left|\dfrac{19n+8}{n^2-10n}\right| < \epsilon$. Only large $n$ matter, so let's assume that $n > 20$. 
Then, $\left|\dfrac{19n+8}{n^2-10n}\right| = \dfrac{19n+8}{n^2-10n} \le \dfrac{19n+n}{\tfrac{1}{2}n^2} = \dfrac{40}{n}$ (you want this to be less than $\epsilon$ for all $n > N$). 
So, try picking $N = \dfrac{40}{\epsilon}$. Then, for all $n > N$, you have $\dfrac{40}{n} < \dfrac{40}{N} = \epsilon$ as you needed. 
However, we assumed that $n > 20$ in order to bound $\left|\dfrac{19n+8}{n^2-10n}\right|$. 
Thus, you need to pick something like $N = \max\left\{\dfrac{40}{\epsilon},20\right\}$ (this ensures that $N > 20$). 
Now that you have found the value of $N$, you can write the proof. 
EDIT: There was a small error in the above inequality, specifically $n^2-10n$ is not smaller than $n^2$. This can be fixed by assuming $n > 20$, which gives $n^2-10n > \tfrac{1}{2}n^2$. Thanks André Nicolas for catching that. 
A: If $n \gt 20$ then $10n\lt \frac{n^2}{2}$ and therefore $n^2-10n \gt \frac{n^2}{2}$.
Also, if $n\gt 20$ then $19n+8\lt 20n$.
So if $n\gt 20$ then our expression is $\lt \frac{20n}{n^2/2}$, which is $\frac{40}{n}$.
Let  $N=\max(20,\lceil 40/\epsilon)\rceil$. If $n\gt N$ then the desired inequality holds.
A: Alternative route to find the limit (so not really an answer to your question):
$$\frac{\left(n+8\right)\left(n+1\right)}{n\left(n-10\right)}=\frac{\left(1+\frac{8}{n}\right)\left(1+\frac{1}{n}\right)}{1\left(1-\frac{10}{n}\right)}$$
Here $\dfrac{1}{n}\rightarrow0$ if $n\rightarrow\infty$, so  $\lim_{n\rightarrow\infty}\frac{\left(n+8\right)\left(n+1\right)}{n\left(n-10\right)}=1$
if $\lim_{x\rightarrow0}\frac{\left(1+8x\right)\left(1+x\right)}{1-10x}=1$.
If $\lim_{x\rightarrow0}f\left(x\right)$ and $\lim_{x\rightarrow0}g\left(x\right)$
both exist with $\lim_{x\rightarrow0}g\left(x\right)\neq0$ then $\lim_{x\rightarrow0}\frac{f\left(x\right)}{g\left(x\right)}$
exists with $\lim_{x\rightarrow0}\frac{f\left(x\right)}{g\left(x\right)}=\frac{\lim_{x\rightarrow0}f\left(x\right)}{\lim_{x\rightarrow0}g\left(x\right)}$.
Apply that here on $f\left(x\right)=\left(1+8x\right)\left(1+x\right)$
and $g\left(x\right)=1-10x$.
