Prove $\lim_{\infty }\frac{\sqrt{n}}{n+1} = 0 $ Please check my proof and point mistake
$$\left |\frac{\sqrt{n}}{n+1}  \right |^{2}<\epsilon^{2} $$
$$ \frac{n}{n^{2}+2n+1}< \epsilon^{2} $$
$$ \frac{1}{\epsilon^{2} }< \frac{n^{2}+2n+1}{n}$$
If we suppose n> N and we choose $N\geq \frac{1}{\epsilon^{2} }$
Then
$$\frac{1}{\epsilon^{2} }< \frac{n^{2}+2n+1}{n}$$
Therefore 
$$\left |s_{n}-0  \right |< \epsilon $$

You are working too hard.
Since
$\dfrac{\sqrt{n}}{n+1}
< \dfrac{\sqrt{n}}{n}
= \dfrac1{\sqrt{n}}
$,
if we can choose
$n$ large enough
so that
$\dfrac1{\sqrt{n}}
< \epsilon
$,
then,
automatically,
$\dfrac{\sqrt{n}}{n+1}
< \epsilon
$.
By elementary algebra,
$\dfrac1{\sqrt{n}}
< \epsilon
\iff
n > \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.
