Question: I am not very familiar to epsilon... But this question belongs to my Sallybus and I want to understand it with the help of epsilon definition.. Kindly help me to solve it

Answer: 
In the first place, it is worth to emphasize the definition of convergence of sequences.

Let $x\_{n}\in X$, where $(X,d)$ is a metric space. We say that $x\_{n}$ converges to $x\_{0}\in X$ iff
\begin{align\*}
(\forall\varepsilon > 0)(\exists n\_{\varepsilon}\in\mathbb{N})\,\,\text{s.t.}\,\,(n\geq n\_{\varepsilon} \Rightarrow d(x\_{n},x\_{0}) < \varepsilon)
\end{align\*}

At the present case, $(X,d) = (\mathbb{R},d)$, $x\_{n} = (n^{2}+1)^{-1}$ and $x\_{0} = 0$, where $d(x,y) = |x - y|$.

Having said that, we can proceed.

To begin with, I would start with noticing that

\begin{align\*}
\frac{1}{n^{2}+1} < \frac{1}{n^{2}} < \varepsilon \Rightarrow n > \frac{1}{\sqrt{\varepsilon}}
\end{align\*}

Thus, for every $\varepsilon > 0$, there corresponds a natural number $n\_{\varepsilon} = \Bigl\lceil\dfrac{1}{\sqrt{\varepsilon}}\Bigr\rceil$ s.t.
\begin{align\*}
n\geq n\_{\varepsilon} \Rightarrow \frac{1}{n^{2}+1} < \varepsilon
\end{align\*}

Hopefully this helps!
Option:

Let $n \ge1;$

$\dfrac{1}{n^2+1} \le\dfrac{1}{n^2} \le \dfrac{1}{n};$

Let $\epsilon >0$ be given.

Archimedian principle:

There exists a $n\_0 \in \mathbb{N}$ s. t.

$n\_0 >1/\epsilon;$

For $n\ge n\_0:$

$\dfrac{1}{n^2+1}\le \dfrac{1}{n}<\dfrac{1}{n\_0} <\epsilon. $