Prove that the sequence $\left\{\frac{n}{n^2+1}\right\}_1^\infty$ converges to 0. 
Prove that the sequence $\left\{\frac{n}{n^2+1}\right\}_1^\infty$ converges to 0.

I'm missing something. I doubt that I'm really doing anything from " $\frac{1}{n+1}$ " on. I set it $< \epsilon$ and then define my lowest n value, $n_0$, by manipulating that inequality. Is that right? If this is the right approach, why? Defining  $\epsilon$ and $n_0$ in terms of each other makes me think I can always make the inequality I make at the end concluding my proof.  
Thank you for your help in explaining this!!
$\left|\frac{n}{n^2+1}-0\right| = \frac{n}{n^2+1} \leq \frac{n}{n^2+n} = \frac{1}{n+1}$
Let $\epsilon>0$ be given. Let $n_0$ be the smallest integer such that $n \geq n_0>\frac{1}{\epsilon}-1$. Equivalently, $\epsilon>\frac{1}{n+1}$.
Thus $\left|\frac{n}{n^2+1}-0\right|<\frac{1}{n+1} < \epsilon$,$\forall n_0 \geq n$, so $\frac{n}{n^2+1}_1^\infty \rightarrow 0.$
 A: Hint:
$$0\leq \frac{n}{n^2+1}\leq \frac{n}{n^2}$$
A: $\underline{\textbf{Proof}}$
Let $\epsilon > 0$ be given.
We need to show $\exists \ N_\epsilon \in \mathbb{N}$ s.t. $|\frac{n}{n^2+1}-0|< \epsilon \implies |\frac{n}{n^2 +1}|<\epsilon$.
We know that $|\frac{n}{n^2 +1}| = \frac{n}{n^2 +1} \forall n\in\mathbb{N}$.
Note also that $\frac{n}{n^2+1}\leq\frac{n}{n^2 +n} = \frac{1}{n+1} \forall n\in\mathbb{N} \ \ \ $. (*)

$\underline{\textbf{Rough Work:}}$
We now have that $\frac{1}{n+1} < \epsilon \implies n+1> \frac{1}{\epsilon}$, since all values are positive.
$\implies n> \frac{1}{\epsilon}-1$

Now, for the given $\epsilon > 0$, using the Archimedean Property, choose $N_\epsilon > \frac{1}{\epsilon} -1$. 
Thus, for $\forall n \geq N_\epsilon$, we have:
$n\geq N_\epsilon > \frac{1}{\epsilon} -1 $
$\implies n+1 \geq N_\epsilon +1 > \frac{1}{\epsilon}$
$\implies \frac{1}{n+1}\leq\frac{1}{N_\epsilon+1} < \epsilon$.
Now, from (*), we have
$\frac{n}{n^2 +1}\leq\frac{1}{N_\epsilon+1}<\epsilon$
$\implies \frac{n}{n^2 +1} < \epsilon$
Thus proving that $\lim_{n\to \infty}(\frac{n}{n^2 +1})=0$.
A: Alternatively: $\frac{n}{n^2+1} = \frac{n+i-i}{(n+i)(n-i)} = \frac{i}{n+i} = \frac{-i}{n-i} \xrightarrow{x \to \infty} 0 $
