Proving limit using the definition , Is my approach correct? I have the following problem: Prove that $$\lim_{n\to \infty} \frac {n^2}{n^2-10n-100}=1$$ using the definition. My approach to the problem was finding $f(n)-1$ which has to be smaller than $\epsilon$. The value of that was $\frac {10n+100}{n^2-10n-100}$, so I said $10n+100<10n$ and $n^2-10n-100 < n^2$, so I find that $n < \frac {10}\epsilon$ : 
The approach shown by my instructor was really complicated, so I want to know if this one is correct. Thank you
 A: Your approach is incorrect based on what lulu's said.
Therefore, for any $\epsilon > 0$, there exists some positive integer $N$ such that
$\displaystyle\left|\frac{n^2}{n^2-10n-100}-1\right|<\epsilon$, for $n>N$
Take $N=10$,
$n>10 => 10n>100 => n^2-10n-10n<n^2-10n-100$
Also note that
$n>10 => 20n > 10n+100$
$\displaystyle\therefore \left|\frac{n^2}{n^2-10n-100}-1\right|=\frac{10n+100}{n^2-10n-100}<\frac{20n}{n^2-10n-10n}=\frac{20}{n-10}<\epsilon$
Hence, take $N=\max(10,{\lceil\frac{20}{\epsilon}+10}\rceil)$
A: You want to show that, given $\varepsilon>0$, there exists $N$ such that, for $n>N$,
$$
\left|\frac{n^2}{n^2-10n-100}-1\right|<\varepsilon
$$
but you do no advance in this direction, unfortunately.

The inequality is equivalent to
$$
-\varepsilon<\frac{10n+100}{n^2-10n-100}<\varepsilon
$$
We can assume $n>20$, so $n^2-10n-100>0$ and the inequality to be solved becomes
$$
n^2-10n-100>\frac{10}{\varepsilon}n+\frac{100}{\varepsilon}
$$
or
$$
n^2-10\left(1+\frac{1}{\varepsilon}\right)n-100\left(1+\frac{1}{\varepsilon}\right)>0
$$
Set for simplicity $1+\frac{1}{\varepsilon}=k$; then the inequality is satisfied for
$$
n>5k+\sqrt{25k^2+100k}
$$
and so you can take
$$
N=\lceil\max(20,5k+\sqrt{25k^2+100k})\rceil
$$
A: Sure, this is a good approach. However, the definition of a limit involves the existence of $N_{\epsilon}$ such that for all $n\geq N_{\epsilon}$, etc. Obtaining $n\color{red}{<}\text{ something }$ should indicate a problem (in this case, since you know that there exists a limit).
You have the sequence $\{f(n)\}_{n}$ with $f(n)=\frac{n^{2}}{n^{2}-10n-100}$. We look for $N_{\epsilon}$ such that for any $n\geq N_{\epsilon}$, the following holds:
$$\vert f(n) -1\vert<\epsilon$$
Solving this, we get:
\begin{align*}
&\left\vert\frac{n^{2}}{n^{2}-10n-100}-1\right\vert\\
&=\left\vert\frac{n^{2}-n^{2}+10n+100}{n^{2}-10n-100}\right\vert\\
&=\left\vert\frac{10n+100}{n^{2}-10n-100}\right\vert\\
&<\epsilon
\end{align*}
Note that $10n+100\geq 0$ for all $n\in\mathbb{N}$. The denominator $n^{2}-10n-100>0$ has two real roots. Take $n\geq $ the bigger one, which is $(5+5\sqrt{5})$. Then for $n\geq 5(1+\sqrt{5})$, the following holds:
$$\left\vert\frac{10n+100}{n^{2}-10n-100}\right\vert=\frac{10n+100}{n^{2}-10n-100}$$
Hence, solving the inequation, we get:
\begin{align*}
0&<\epsilon n^{2}-10\epsilon n -100\epsilon -10n -100\\
&<n^{2}\epsilon-10(\epsilon+1)n-100(\epsilon+1)
\end{align*}
The discriminant is $\Delta=10(\epsilon+1)^{2}+400\epsilon(\epsilon+1)=(\epsilon+1)(10+410\epsilon)$ hence, the bigger root is
$$n_{+}=\frac{10(\epsilon+1)+\sqrt{(\epsilon+1)(10+410\epsilon)}}{2\epsilon}$$
Now, defining:
$$N_{\epsilon}=\left\lceil \max\left\{5(1+\sqrt{5})\,,\,\frac{10(\epsilon+1)+\sqrt{(\epsilon+1)(10+410\epsilon)}}{2\epsilon}\right\}\right\rceil$$
you have proven that whatever is $\epsilon>0$, you can find a $N_{\epsilon}$ such that for any $n\geq N_{\epsilon}$, we have:
$$\vert f(n)-1\vert<\epsilon$$
