<s>[INST] How prove $\lim x_n = \sqrt{\frac{a}{b}}$ if $b \neq 0$ for $x_1=c$ and $x_{n+1}=\frac{x_n^2 + \frac{a}{b}}{2x_n}$? Let $x_1=c$ and $x_{n+1}=\frac{x_n^2 + \frac{a}{b}}{2x_n}$. How prove  $\lim x_n = \sqrt{\frac{a}{b}}$ if $b \neq 0$ ?
 [/INST] Let $r=\frac{a}{b}$ and assume that $r>0$ and $c>0$. To prove existence of the limit, rewrite the equation as follows:
$$
2x_{n+1}=x_n + \frac{r}{x_n}\quad (\star)
$$
Since $x_1=c>0$ and $r>0$, it follows by induction that all $x_n$ are positive. Applying the AM-GM Inequality, we see that $x_{n+1}\geq \sqrt{r}$ (for $n\geq 1$). Squaring both sides of $(\star)$ yields
$$
\begin{align*}
4x_{n+1}^2&=x_n^2 + 2r + \frac{r^2}{x_n^2}\\
\implies 4x_{n+1}^2-4r&=\left(x_n-\frac{r}{x_n}\right)^2\\
\implies \left|\frac{x_{n+1}^2-r}{x_n^2-r}\right|&=\frac{\left|1-\frac{r}{x_n^2}\right|}{4}.
\end{align*}
$$
Since $x_n\geq \sqrt{r}$ for $n\geq 2$, it follows that $\left|1-\frac{r}{x_n^2}\right|\leq 1$ for $n\geq 2$. Consequently
$$
\left|x_{n+1}^2-r\right|\leq \left(\frac{1}{4}\right)\cdots \left(\frac{1}{4}\right)\left(\frac{\left|1-\frac{r}{c^2}\right|}{4}\right)\left|c^2-r\right|=\frac{C}{4^n},
$$
for some positive constant $C$.
Taking the limit as $n\to\infty$, it follows that $\lim_{n\to\infty} x_{n+1}^2=r$, and therefore $\lim_{n\to\infty}x_n=\sqrt{r}$ as desired.
</s>