Prove $\lim_{\infty }\frac{\sqrt{n}}{n+1} = 0 $ Please check my proof and point mistake

$$\left |\frac{\sqrt{n}}{n+1}  \right |^{2}&lt;\epsilon^{2} $$

$$ \frac{n}{n^{2}+2n+1}&lt; \epsilon^{2} $$

$$ \frac{1}{\epsilon^{2} }&lt; \frac{n^{2}+2n+1}{n}$$

If we suppose n> N and we choose $N\geq \frac{1}{\epsilon^{2} }$

Then

$$\frac{1}{\epsilon^{2} }&lt; \frac{n^{2}+2n+1}{n}$$

Therefore 

$$\left |s_{n}-0  \right |&lt; \epsilon $$
You are working too hard.

Since
$\dfrac{\sqrt{n}}{n+1}
&lt; \dfrac{\sqrt{n}}{n}
= \dfrac1{\sqrt{n}}
$,
if we can choose
$n$ large enough
so that
$\dfrac1{\sqrt{n}}
&lt; \epsilon
$,
then,
automatically,
$\dfrac{\sqrt{n}}{n+1}
&lt; \epsilon
$.

By elementary algebra,
$\dfrac1{\sqrt{n}}
&lt; \epsilon
\iff
n &gt; \dfrac1{\epsilon^2}
$
and we are done.

This is an example
of a general principle
when proving limits:
Lower order terms can
almost always be disregarded.

When this is done,
the resulting value of $n$
may not be the best,
but, if the goal is to show
that the limit exists,
that does not matter.
You need $N\le \frac{1}{\epsilon ^2}$ and $n&lt;N$ which gives $\frac{n^2+2n+1}{n} &gt; \frac{1}{\epsilon ^2}\ge N&gt; n$ for $n&gt;N$, a contradiction.

(Even if this works, it requires $n&gt;0$ for line 2-3 and the argument above.)