Proof verification that $ \lim_{n\to \infty} \frac{n^2+n-1}{n^2 + 2n +2}=1$ EDIT: I've had some problems uploading this question today as I initially used the mobile verision, hence the quite absurd first proof if you saw it. Here is the full one:
We do this using the epsilon-N definition of the limit of :
$$\forall \varepsilon>0 ,\hspace{1mm} \exists N>0 \hspace{1mm}\text{s.t}\hspace{2mm} n\geq N \implies |a_{n}-L|<\varepsilon$$
Now,
\begin{align}
|a_{n}-L| & = \left| \frac{n^2+n-1}{n^2 + 2n +2}-1 \right| \\ &=\left| \frac{-n-3}{n^2 + 2n +2} \right|
\end{align}
We now consider the definition of $|x|$ for q quick moment, where we have that:
$$|x|=\left\{
    \begin{aligned}
        &x \hspace{2mm}\text{if}  \hspace{2mm}x\geq0\\
        -&x \hspace{2mm}\text{if}  \hspace{2mm}x<0
    \end{aligned} \right\}
$$
Since we can split the absolute value into the numerator and denominator separately, we have that:
$$|-n-3|=n+3 \hspace{2mm} \text{if} \hspace{2mm} -n-3<0 \iff -3<n$$
However, in our proof, we require $N>0$, thereby $n>0$, meaning that $|-n-3|=n+3$. A similar argument shows that $|n^2 +2n +2|=n^2+2n+2$ for $n>0$. Thereby, for $n>0$:
\begin{align}
|a_{n}-L| &=\frac{n+3}{n^2 + 2n +2} \\
&\leq \frac{n+3}{n^2+2n}
\end{align}
For $n\geq 3$,
\begin{align}|a_n-L| &\leq \frac{n+3}{n^2+2n} \\
&\leq \frac{2n}{n^2+2n} \\
&= \frac{2}{n+2} \\
&\leq \frac{2}{n} < \varepsilon
\end{align}
Thus, to complete our proof, we choose $N=\max\{{\frac{2}{\varepsilon},3}\}$.
Does this look good?
 A: HINT
As a suggestion, you do not need to split into two cases. Suppose that $n\geq n_{\varepsilon}$. Then one gets that
\begin{align*}
\left|\frac{n^{2} + n - 1}{n^{2} + 2n + 2} - 1\right| & = \frac{n + 3}{n^{2} + 2n + 2} = \frac{n + 3}{(n + 1)^{2} + 1} \leq \frac{3n + 3}{(n + 1)^{2}} \leq \frac{3}{n} \leq \frac{3}{n_{\varepsilon}}
\end{align*}
Can you take it from here?
A: Do
$\frac{n^2+n-1}{n^2+2n+2}=\frac{1+\frac{1}{n}-\frac{1}{n^2}}{1+\frac{2}{n}+\frac{2}{n^2}}$
Make use of the fact that this is each $1$ plus something that is getting smaller and smaller the bigger $n$ gets.
$\frac{1}{1+x}\approx 1-x$
in first order for small $x$.
So the division goes over into a simple binomial multiplication for larger $n$.
$(1+\frac{1}{n}-\frac{1}{n^2})(1-\frac{2}{n}-\frac{2}{n^2})$
The leading term is already the desired $1$ and the work is done without conducting the multiplication.
One can do
$(1+\frac{1}{n})(1-\frac{2}{n})=1-\frac{1}{n}-\frac{2}{n^2}$
To show that the approximation towards one is for large natural $n$ dominated by $-\frac{1}{n}$.
The trick with $\frac{1}{1+x}\approx 1-x$ is usually or at least often introduced in courses facing this kind of tasks. An appropriate reference will be a plus in the answer.
Representate with a value table:

or the corresponding plot:

