Use the formal definition of a limit for sequences to prove the limit of a sequence . Where Did I Go Wrong? The formal definition of limit of a sequence says that  for every $\epsilon>0$, there is a number $N>0$ such that if $n > N$  it is true that $|(a_n) - L| <  ε$
In this particular case I have to demonstrate by the definition that this limit exist:
$$\lim \limits_{n \to \infty}\frac{(1-4n-7n^2)}{1+2n+n^2}=-7$$
This is what I did :
$\|\frac{1-4n-7n^2}{1+2n+n^2} + 7\|$ = $\|\frac{10n+8}{(n+1)^2}\|$
In the definition, we have that $N > 0$, and $n>N$, this implies that
$$\|\frac{10n+8}{(n+1)^2}\| = \frac{10n+8}{(n+1)^2}$$
I then used the fact that :
$\frac{10n+8}{(n+1)^2}$ < $10n+8$
$10n+8 < ε$  if and only if  $n < \frac{ε - 8}{10}$
I dont know what to do after that. Also, I don't think that makes sense, In these types of proofs, my teacher always had n > in the end and not vice versa. Also, it doesn't make sense that n needs to be very small since we are observing to what value the sequence converges to when n is VERY LARGE  I don't know where I went wrong and how to do it otherwise.
 A: You were on the right track.  You wrote
$$\left|\frac{1-4n-7n^2}{(n+1)^2}-(-7)\right|=\left|\frac{10n+8}{(n+1)^2}\right|\tag1$$
But while the right-hand side of $(1)$ is less than $10n+8$, this doesn't help here.  Instead, you want to show is that for any given $\varepsilon>0$, you can find a number $N$ so that for any $n>N$, $\left|\frac{10n+8}{(n+1)^2}\right|<\varepsilon$.
So, let's pause and see what we have.  Notice that as $n\to \infty$, the numerator of the right-hand side of $(1)$ grows like $10n$ and the denominator grows like $n^2$. So, we expect the ratio to behave like $\frac{10}{n}$.
Now, notice that $(n+1)^2>n^2$ so that $\frac1{(n+1)^2}<\frac1{n^2}$.  And notice that for $n\ge 1$, $10n+8<20n$.  Putting these inequalities together we find that for any $\varepsilon>0$
$$\left|\frac{10n+8}{(n+1)^2}\right|<\frac{20}{n}<\varepsilon$$
whenever $n>\frac{20}{\varepsilon}$.  So, given an $\varepsilon>0$, we have found a number $N=\frac{20}{\varepsilon}$ such that whenever $n>N$, the right-hand side of $(1)$ is less than $\varepsilon$.  And we are done!
