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.

Since the given sequence is positive for all $n\geq 1$, we can drop the absolute value signs. 
Consider the inequality:
\begin{equation*}
\frac{n+1}{n^2+1}<\epsilon
\end{equation*}
This is a messy inequality, as solving for $n$ would be rather difficult. Instead we shall find an upper bound for the numerator $(n+1)$ and a lower bound on the denominator $(n^2+1)$ so that we can construct a new sequence $f_{n}$ such that $$\frac{n+1}{n^{2}+1}<f_{n}<\epsilon.$$
Since $n+1$ behaves like $n$ at very large numbers, we shall try to find a value for $b$ such that $bn>n+1$. And since $n^{2}+1$ behaves like $n^{2}$ for very large numbers, we shall try to find a value for $c$ such that $cn^{2}<n^{2}+1$. $n+1<n+n=2n$ and $n^{2}+1>\frac{1}{2}n^{2}$ for all $n\geq 1$.
Now, consider the new inequality:
\begin{align*}
\frac{n+1}{n^{2}+1}<\frac{2n}{\frac{1}{2}n^{2}}<\frac{4}{n}<\epsilon
\end{align*}
We have 
\begin{equation*}
\frac{4}{n}<\epsilon\iff \frac{4}{\epsilon}<n
\end{equation*}
Let $\displaystyle N(\epsilon)=\lfloor4/\epsilon\rfloor$. Therefore, $\forall \epsilon>0,\ \exists N(\epsilon)=\lfloor4/\epsilon\rfloor\ni\forall n>N(\epsilon)$
\begin{align*}
\frac{n+1}{n^{2}+1}&<\frac{4}{n}<\frac{4}{N(\epsilon)+1}<\frac{4}{4/\epsilon}=\epsilon
\end{align*}
