Answer
$\sqrt a$
Work Step by Step
Let us consider that $a_{n+1} \approx a_n =l$
We have: $\lim\limits_{n \to \infty}x_{n+1}=\dfrac{1}{2} \lim\limits_{n \to \infty}(x_{n}+\dfrac{a}{x_n})\\=\dfrac{1}{2}(l+\dfrac{a}{l})$
This yields: $2l=l+\dfrac{a}{l}\\ \implies l^2=a \\ \implies l=\pm \sqrt a$
But the limit does not attain negative values for a positive sequence, so $l=-\sqrt a$ must be neglected.
So, our limit is $l=\sqrt a$.
