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 
$$\left |s_{n}-0  \right |< \epsilon $$
 A: 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.
A: Minor stuff: You are missing some formal notation:

$$\color{red}{\left|\frac{\sqrt{n}}{n+1}\right|<\frac{1}{\epsilon}}$$
$$\color{red}{\because}\left |\frac{\sqrt{n}}{n+1}  \right |^{2}<\epsilon^{2} $$
$$\color{red}{\because} \frac{n}{n^{2}+2n+1}< \epsilon^{2} $$
$$\color{red}{\because} \frac{1}{\epsilon^{2} }< \frac{n^{2}+2n+1}{n}$$

This next sentence is vague. Why are you supposing this? You should have a definition you are working form which should maybe be at the start. Something like:
$$\color{red}{\text{The sequence }s_n\text{ converges to }L\text{ if given }\epsilon>0\exists N\text{ such that }|s_n-L|<\epsilon\forall n>N}$$

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
$$\left |s_{n}-0  \right |< \epsilon $$

Given the style you are using I would have finished off something like this:
$$\color{red}{\because n<\epsilon^2(n^2+2n+1)}$$
$$\color{red}{\because 0<\epsilon^2n^2-(1-\epsilon^2)n+\epsilon^2}$$
$$\color{red}{\text{Taking the positive root (as }n>0\text{) :}}$$
$$\color{red}{\because n>\frac{1-\epsilon^2+\sqrt{(1-\epsilon^2)^2-4\times\epsilon^2\times\epsilon^2}}{2\epsilon^2}}$$
$$\color{red}{\because \text{Let }N=\frac{1-\epsilon^2+\sqrt{(1-\epsilon^2)^2-4\times\epsilon^2\times\epsilon^2}}{2\epsilon^2}}$$
$$\color{red}{\text{So if }n>N\text{ then }\left|s_n-0\right|<\epsilon}$$
A: Rearranging gets $sqrt (n/(n+1)^2)$ . The sqrt goes to 0 when the function value goes to 0. Thus we consider $n/(n+1)^2$ which goes to 0 when its reciprocal goes to infinity. It is obvious the reciprocal goes to infinity when $n$ goes to infinity because it is just $n+1/n$    .
