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.
 A: 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*}
A: Hint: Show that $0<\frac{n+1}{n^2+1}<\frac{1}{n-1}$.
A: 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}
A: Fairly crude  bounds or estimates can be very useful in simplifying  problems about limits. For example if $n\geq 1$ then $$0<\frac {n+1}{n^2+1}=\frac {n(1+n^{-1})}{n^2(1+n^{-2})}=\left(\frac {1}{n}\right)\cdot \frac {1+n^{-1}}{1+n^{-2}}\leq \left( \frac {1}{n}\right) \cdot\frac {2}{1+n^{-2}}< \left(\frac {1}{n}\right)\cdot\frac {2}{1}=\frac {2}{n}.$$ So if $\epsilon >0$ and $n\geq \max (1,\frac {2}{\epsilon})$ then $|\frac {n+1}{n^2+1}|<\epsilon.$
A: Let $\epsilon <1$ then $$\frac{n+1}{n^2+1}<\epsilon \Leftrightarrow \epsilon n^2-n-(1-\epsilon)>0$$
You can solve this inequeality like a quadratic polinomial and you can find that for $$\forall n\geqslant n_0=[\frac{1+2\sqrt{\epsilon(1-\epsilon)}}{2}]+1 \Rightarrow \frac{n+1}{n^2+1}<\epsilon$$ forall $\epsilon<1$.
Thus we found an appropraite $n_0 \in \mathbb{N}$,which depends on $\epsilon$, such that $\frac{n+1}{n^2+1}<\epsilon,\forall n\geqslant n_0$
This is a way to prove it
but indeed it is easier to find a simpler sequence $b_n$ such that $\frac{n+1}{n^2+1} \leqslant b_n<\epsilon$.
A: Examining $\frac{n+1}{n^2+1}$ you realize that as $n$ grows it will start 'acting like' $1/n$. You know the quadratic formula, and after a bit of tinkering you guess that the key is analyzing
$\tag 1 \frac{n+1}{n^2+1} \le \frac{2}{n}$
The inequality (1) is equivalent to
$\tag 2 n^2 - n + 2 \ge 0$
The discriminant for this quadratic expression is negative, so it is always true.
Let the $\varepsilon \gt 0$ challenge be presented. Then $2/n \lt \varepsilon$ iff $n \gt 2/\varepsilon$.
Select any natural number $N$ greater than $2/\varepsilon$.
We have shown that if $n \ge N$, $\frac{n+1}{n^2+1} \lt \varepsilon$, and the proof of convergence had been demonstrated.
