Show the limit of the sequence is $\sqrt{\alpha}$ Given: $\sqrt{\alpha}<x_1, \alpha >0, x_{n+1}=\frac{1}{2}\left(x_n+\frac{\alpha}{x_n}\right)$. Show $\lim_{n\to \infty}x_n=\sqrt{\alpha}.$
I have already shown the this is a monotonic decreasing function. Intuitively, I see in the end I get $\frac{1}{2}\left(\sqrt{\alpha}+\sqrt{\alpha}\right)$ but I'm not sure if I should keep going using $\lim$ notation or if I should switch to the definition of converging sequence.

Rewrite $$ x_{n+1}=\frac{1}{2}\left(x_n+\frac{\alpha}{x_n}\right)$$
$$ x_{n+1}= x_{n}-\Big[x_n-\frac{1}{2}\left(x_n+\frac{\alpha}{x_n}\right)\Big]$$ that is to say
$$ x_{n+1}= x_{n}-\frac {x_n^2-\alpha}{2x_n}$$ and recognize the Newton iterative scheme for finding the zero of function $f(x)=x^2-\alpha$.
