Sequence limits using epsilon I just wanted to find out a few examples of how to use the epsilon characterisation to for limits. One example should help me get my head around these and do other questions.
$A_n=2/(n^2+\sqrt{3})$ prove using epsilon characterisation that $A_n$ has limit $0$ as $n$ tends to infinity

EDITED WITH MORE DETAILS

Definition. We say that a sequence of numbers $a_1,a_2,\dots$ converges to a number $a$ if for each $\varepsilon>0$ you can find an $n_0\in\mathbf N$ such that for every $n\geq n_0$ it holds:
$$|a_n-a|<\varepsilon.$$

In other words, the sequence converges to $a$ if eventually the terms are "as close as you want" to $a$. 
In your example you have to show that for each $\varepsilon>0$ there exists an $n_0\in\mathbf N$ such that if $n\geq n_0$ then $|A_n-0|<\varepsilon$ (because $A_n$ converges to zero.) 
We now note that
$$|A_n|=\frac{2}{n^2+\sqrt{3}}\leq\frac 2{n^2}.$$
We do that because it is simpler to work with $\frac 2{n^2}$. Now, since for any $n\geq n_0$ it holds that 
$$\frac2{n^2}\leq\frac2{n_0^2},$$
we bound $|A_n|$ ($n\geq n_0$) as follows:
$$|A_n|\leq\frac2{n^2}\leq\frac2{n_0^2}.$$
Now suppose that $\frac2{n_0^2}<\varepsilon$. In that case, we would obtain that 
$$|A_n|\leq\frac2{n^2}\leq\frac2{n_0^2}<\varepsilon,$$
and that is exactly what we want! But, do the condition $\frac2{n_0^2}<\varepsilon$ ever holds? Yes, but only if $n_0>\sqrt{\varepsilon/2}$. But there is no problem, since we are free to choose the $n_0$. So let $n_0>\sqrt{\varepsilon/2}$ (for example, $n_0=[\sqrt{\varepsilon/2}]+1$, where $[\cdot]$ is the floor function.) Then 
$$|A_n|\leq\frac2{n^2}\leq\frac2{n_0^2}<\frac{2}{(\sqrt{\varepsilon/2})^2}=\varepsilon.$$
By our definition, this proves that $A_n$ converges to $0$. 
Now we consider your other example, $B_n=\sqrt{5}+\frac{2}{n^2+\sqrt{3}}$. We will show that $B_n$ converges to $\sqrt{5}$. Read again the definition. What we have to do? To study the difference $|B_n-\sqrt{5}|$. Let's do it: 
$$|B_n-\sqrt{5}|=\left|\sqrt{5}+\frac{2}{n^2+\sqrt{3}}-\sqrt{5}\right|=
\left|\frac{2}{n^2+\sqrt{3}}\right|.$$
The term $\sqrt{5}$ has disappeared! So if we want to prove that $B_n$ tends to $\sqrt{5}$, we only have to find $n_0$ such that $|\frac{2}{n^2+\sqrt{3}}|<\varepsilon$. Is it possible? Indeed, we have just done it above. 
