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.

Since this hasn't been closed as duplicate,
here's my answer again.
A more direct proof.
$s_{n+1}
=\dfrac12(s_n+\dfrac{a}{s_n})
$.
$\begin{array}\\
s_{n+1}^2
&=\dfrac14(a_n^2+2a+\dfrac{a^2}{s_n^2})\\
&=a+\dfrac14(s_n^2-2a+\dfrac{a^2}{s_n^2})\\
&=a+\dfrac14(s_n-\dfrac{a}{s_n})^2\\
&\gt a\\
\end{array}
$
so
$s_n > \sqrt{a}
$.
$\begin{array}\\
s_{n+1}-\sqrt{a}
&=\dfrac12(s_n-2\sqrt{a}+\dfrac{a}{s_n})\\
&=\dfrac12(\sqrt{s_n}-\dfrac{\sqrt{a}}{\sqrt{s_n}})^2\\
&=\dfrac1{2s_n}(s_n-\sqrt{a})^2\\
\dfrac{s_{n+1}-\sqrt{a}}{s_n-\sqrt{a}}
&=\dfrac1{2s_n}(s_n-\sqrt{a})\\
&\lt \dfrac12\\
\end{array}
$
so $s_n$ converges to
$\sqrt{a}$.
