How to prove the limit of a sequence using "$\epsilon-N$" I think I have a proper understanding of the general procedure, but I'm having difficulty manipulating my inequality so that I can isolate $n$ by itself. Sadly I wasn't given many examples to model my answer on.

Prove that $\displaystyle\lim_{n\to\infty}\frac{n+1}{n^2+1}=0$

So I'm given $L=0$. I then look at the inequality
$$\left| \frac{n+1}{n^2+1}-0\right|<\epsilon$$
but I have no idea how to isolate $n$. The best I can come up with, which may be the right idea, is to use another function $f$ such that
$$\left|\frac{n+1}{n^2+1}\right|<f<\epsilon$$
and then work with that. But my idea of using $f=\lvert n+1\rvert$ seems to have me a bit stuck too.

The $\epsilon-N$ definition to $\lim_{n\to \infty}a_n=L$ is
$$
\big(
\forall \epsilon>0 
\big)
\big(
\exists N_{\epsilon}\in \mathbb{N}
\big)
\big[ 
(
\forall n\in \mathbb{N} 
)
( 
n>N_{\epsilon}
)
\implies
(
|a_n-L|<\epsilon
)
\big]
$$
That is, given an arbitrary, but fixed, $\epsilon>0$, we must find a candidate for $N_\epsilon\in\mathbb{N}$ with the property that 
$$
(\forall n\in \mathbb{N} 
)
( 
n>N_{\epsilon}
)
\implies
(
|a_n-L|<\epsilon
)
$$
The number $N_\epsilon$ also depends on the limit $L$ and the sequence itself as well. In this case, $L=0$ and 
$$
a_n= \frac{n+1}{n^2+1}.
$$
To find an $N_\epsilon$ candidate with the required property we must manipulate the inequality
$$
0-\epsilon< \frac{n+1}{n^2+1}< 0 +\epsilon.
$$
The inequality of the left is valid for all $\epsilon> 0$. Focus on the inequality of the right
$$
\frac{n+1}{n^2+1}< \epsilon.
$$
Note that
$
\frac{n+1}{n^2+1}
<
\frac{n+1}{n^2-1}
=
\frac{1}{n-1}
$ for $n\geq 2$. Once $\frac{1}{n-1}<\epsilon$ implies $\frac{n+1}{n^2+1}<\epsilon$, set 
$$
N_\epsilon =\min\left\{n\in\mathbb{N}: n>\frac{1}{\epsilon}+1 \mbox{ and } n\geq 2 \right\}
$$
In fact, 
\begin{align}
n>N_\epsilon \implies & n>\frac{1}{\epsilon}+1 \mbox{ and } n\geq 2
\\
\implies & n>\frac{1}{\epsilon}+1 \mbox{ and } n\geq 2
\\
\implies & \frac{1}{\epsilon}<n-1 \mbox{ and } n\geq 2
\\
\implies & n-1>\frac{1}{\epsilon} \mbox{ and } n\geq 2
\\
\implies & \frac{1}{n-1}<\epsilon \mbox{ and } n\geq 2
\\
\implies & \frac{n+1}{n+1}\cdot\frac{1}{n-1}<\epsilon
\\
\implies & \frac{n+1}{n^2-1}<\epsilon
\\
\implies & \frac{n+1}{n^2+1}<\epsilon 
\\
\implies & -\epsilon< \frac{n+1}{n^2+1}< \epsilon.
\\
\implies & \left|\frac{n+1}{n^2+1}\right|< \epsilon.
\end{align}
