<s>[INST] How can I show the convergence and find limit. 
Let $a_1=0$ and $$a_{n+1}=\frac{a_n^2+3}{2(a_n+1)}~,~n \geq 1$$ . Show that $\{\ a_n \}$ converges  and find its limit.

Trial: Here $$\begin{align} a_1 &=0\\ \text{and}~ a_2 &=\frac{3}{2} \\ \text{and}~ a_3 &=\frac{21}{20} \end{align}$$ . Here I find no relationship between $a_n ~\text{and}~ a_{n+1}$. Please help.
 [/INST] $a_1=0 \Rightarrow a_2= \frac{0^{2}+3}{2(0+1)}=\frac{3}{2}$. Note that $\frac{3}{2}\geqslant\frac{3}{2}>1$.
Next, $a_n>1\Rightarrow \exists k>0$ such that $a_n = 1+k$.
Also, if $k\leqslant \frac{1}{2}$, then
$a_{n+1}=\frac{(1+k)^{2}+3}{2((1+k)+1)}=\frac{4+2k+k^{2}}{4+2k}=1+\frac{k^{2}}{4+2k}(*)<1+\frac{k^{2}}{4}+\frac{k^{2}}{2k}=1+\frac{k}{2}(1+\frac{k}{2})$
$\leqslant 1+\frac{k}{2}(1+\frac{1}{4})=1+\frac{k}{2}(\frac{5}{4})=1+ \frac{5}{8}k<1+k=a_n.$
I noticed the inequality at $(*)$ works only after I deleted this answer, because at first it was a genuine schoolboy error, but it actually is true and therefore I can luckily salvage this answer. Why is $(*)$ true? Because $a>0,b>0,c>0\Rightarrow a<2a=\frac{ab}{b}+\frac{ac}{c}<\frac{a(b+c)}{b}+\frac{a(b+c)}{c} \ $. Now divide both sides by $(b+c)$: $\quad \frac{a}{b+c}<\frac{a}{b}+\frac{a}{c}. \ $ In fact, we could make the restriction of k above even stronger: $1+\frac{k^{2}}{4+2k}<1+ \frac{5}{16}k \ $ because $\quad \frac{a}{b+c}<\frac{1}{2}(\frac{a}{b}+\frac{a}{c}), \ $ however I have already used $\frac{5}{8}$ throughout the rest of the proof and it suffices, so let's stick with that.
From this, we see that:
$$\bbox[yellow]{(1)\qquad 1<a_{n+1}<1+\frac{5}{8}k<1+k=a_n\leqslant\frac{3}{2}\qquad for\ all \ n\geqslant 2.}\ $$
From $(1)$ we deduce: {$a_n$} is monotonically decreasing and converges to some limit $L$ where $\frac{3}{2}\geqslant L\geqslant1$, and also $$\bbox[yellow]{(2)\quad\qquad\qquad\qquad\qquad a_{n+1}-1<\frac{5}{8}(a_n-1) \qquad for\ all \ n\geqslant 2,}\ $$
because $k=a_n-1.$
$(2)$ says that the distance between $a_{n+1}$ and 1 is $<\frac{5}{8}$ of the distance between $a_n$ and 1.
Thus $$\bbox[yellow]{(3)\quad\qquad\qquad\qquad\qquad 1<a_n\leqslant 1+\frac{1}{2}(\frac{5}{8})^{n-2} \qquad for\ all \ n\geqslant 2.}\ $$
At this stage, I would invoke the Sandwich Theorem. Our sequence {$a_n$} is "squeezed" or "sandwiched" by the two sequences {$b_n$} = {$1$} and {$c_n$} = {$1+\frac{1}{2}(\frac{5}{8})^{n-2}$}, both of which converge to $1$. Therefore, by the Sandwich Theorem for sequences, $\ L=1$.
Instead of invoking the Sandwich Theorem, you could also go the "formal route".
Given ${\varepsilon}>0,$ no matter how small, we need to find $N$, in terms of ${\varepsilon}$, such that $a_n-1<{\varepsilon} \ $ for all $n \geqslant N$
$a_n-1<{\varepsilon} \iff a_n \leqslant 1+\frac{1}{2}(\frac{5}{8})^{n-2}<1+{\varepsilon} \iff (\frac{5}{8})^{n-2}<2{\varepsilon} \iff (n-2)log(\frac{5}{8}) <log(2{\varepsilon}) \iff n-2> \frac{log(2{\varepsilon})}{log(5/8)} \iff$
$$\bbox[yellow]{n>\frac{log(2{\varepsilon})}{log(5/8)}+2}\ , $$
and so we have found $N({\varepsilon})$, the least integer greater than $\frac{log(2{\varepsilon})}{log(5/8)}+2$, which guarantees $a_n-1<{\varepsilon}$ for all $n\geqslant N.$
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